u-Substitution - Business Calculus

1. The rate of change of a company's revenue is modeled by $R'(t) = (t+1)e^{0.2t^2+0.4t}$ thousand dollars per year, where $t$ is the number of years from the present. Find the total revenue generated during the first 2 years (from $t=0$ to $t=2$).
   Answer: $2.5(e^{1.6} - 1)$ When you see a rate of change function and need to find total change over an interval, you're dealing with the Fundamental Theorem of Calculus. Since $R'(t)$ represents the rate of revenue change, the total revenue over time is found by integrating: $\int_0^2 R'(t) \, dt = \int_0^2 (t+1)e^{0.2t^2+0.4t} \, dt$ To solve this integral, notice that the exponent $0.2t^2 + 0.4t$ has a derivative of $0.4t + 0.4 = 0.4(t+1)$. This suggests substitution: let $u = 0.2t^2 + 0.4t$, so $du = 0.4(t+1)dt$, which means $(t+1)dt = \frac{1}{0.4}du = 2.5du$. When $t = 0$: $u = 0$ When $t = 2$: $u = 0.2(4) + 0.4(2) = 0.8 + 0.8 = 1.6$ The integral becomes: $\int_0^{1.6} 2.5e^u \, du = 2.5[e^u]_0^{1.6} = 2.5(e^{1.6} - e^0) = 2.5(e^{1.6} - 1)$ Answer A uses the wrong coefficient (0.4 instead of 2.5). Answer C omits the coefficient entirely. Answer D forgets to subtract 1, missing the evaluation at the lower limit of integration. Study tip: When you see $(t+1)$ multiplied by an exponential where the exponent's derivative involves $(t+1)$, immediately think u-substitution. Always remember that definite integrals require evaluating at both limits and subtracting.
2. What is the total increase in profit when production increases from 0 to 4 units?
   Answer: $18 \ln(\frac{5}{3})$ The total increase in profit is the definite integral of the marginal profit function from $x=0$ to $x=4$: $\int_{0}^{4} \frac{18x}{x^2+9} dx$ Let $u = x^2+9$. Then $du = 2x \, dx$. The term $18x \, dx$ can be written as $9(2x \, dx) = 9 \, du$. Change the limits of integration: when $x=0$, $u=0^2+9=9$; when $x=4$, $u=4^2+9=25$. The integral becomes: $\int_{9}^{25} \frac{9}{u} du = 9 [\ln|u|]_{9}^{25}$ Evaluate at the new bounds: $9(\ln(25) - \ln(9)) = 9\ln(\frac{25}{9}) = 9\ln((\frac{5}{3})^2) = 9 \cdot 2\ln(\frac{5}{3}) = 18\ln(\frac{5}{3})$ Distractor B results from forgetting the factor of 9 from the substitution. Distractor C arises from the incorrect logarithm property $\ln(a) - \ln(b) = \ln(a-b)$. Distractor D comes from incorrectly integrating $\frac{1}{u}$ as $-\frac{1}{u^2}$ and evaluating.
3. A company's marginal cost function is given by $C'(x) = 6x(2x^2 + 5)^3$ dollars per unit. What substitution should be used to find the total cost function $C(x)$?
   Answer: Let $u = 2x^2 + 5$, so $du = 4x\,dx$ and $x\,dx = \frac{1}{4}du$ To integrate $\int 6x(2x^2 + 5)^3\,dx$, we need to identify that $x\,dx$ appears as a factor and $(2x^2 + 5)$ is raised to a power. Setting $u = 2x^2 + 5$ gives $du = 4x\,dx$, so $x\,dx = \frac{1}{4}du$. This transforms the integral to $\int 6 \cdot \frac{1}{4} u^3\,du = \frac{3}{2}\int u^3\,du$. Choice B incorrectly substitutes for the coefficient. Choice C substitutes for the entire expression raised to the third power, making the integral more complex. Choice D only substitutes part of the expression inside the parentheses.
4. To evaluate $\int \frac{x}{(3x^2 + 7)^4}\,dx$, a substitution $u = 3x^2 + 7$ is used. After substitution, the integral becomes $\int \frac{1}{6u^4}\,du$. What is the final answer?
   Answer: $-\frac{1}{18u^3} + C = -\frac{1}{18(3x^2 + 7)^3} + C$ When you encounter an integral with a complex expression raised to a power, substitution is often the key technique. Here, the substitution $u = 3x^2 + 7$ transforms the messy original integral into the simpler form $\int \frac{1}{6u^4}\,du$. To evaluate $\int \frac{1}{6u^4}\,du$, rewrite it as $\frac{1}{6}\int u^{-4}\,du$. Using the power rule for integration, $\int u^n\,du = \frac{u^{n+1}}{n+1} + C$ (when $n \neq -1$), you get: $\frac{1}{6} \cdot \frac{u^{-4+1}}{-4+1} + C = \frac{1}{6} \cdot \frac{u^{-3}}{-3} + C = \frac{1}{6} \cdot \frac{-1}{3u^3} + C = -\frac{1}{18u^3} + C$ Substituting back $u = 3x^2 + 7$ gives $-\frac{1}{18(3x^2 + 7)^3} + C$. Answer A has the wrong sign—it's positive instead of negative, missing the negative that comes from integrating $u^{-4}$. Answer B correctly identifies the negative sign but uses the wrong coefficient; it shows $-\frac{1}{6}$ instead of $-\frac{1}{18}$, forgetting to multiply by the $\frac{1}{3}$ from the power rule. Answer D has the wrong exponent entirely—it shows $u^{-5}$ instead of $u^{-3}$, indicating confusion about how the power rule works. Study tip: When integrating $u^n$ where $n < 0$, the result will have exponent $n+1$ (which is less negative) and be divided by $n+1$ (which is negative when $n < -1$). Always double-check your arithmetic with the coefficients and signs.
5. A student attempts to evaluate $\int 2x\cos(x^2 + 1)\,dx$ and writes: "Let $u = x^2 + 1$, so $du = 2x\,dx$. The integral becomes $\int \cos(u)\,du = \sin(u) + C = \sin(x^2 + 1) + C$." What can be concluded about this solution?
   Answer: The solution is completely correct and well-executed throughout all steps. When you encounter an integral with a function and its derivative present, substitution is often the right approach. This problem tests your ability to recognize and execute the substitution method correctly. Let's verify each step of the student's work. The substitution $u = x^2 + 1$ is well-chosen because the derivative of $x^2 + 1$ is $2x$, which appears as a factor in the integrand. When $u = x^2 + 1$, then $\frac{du}{dx} = 2x$, so $du = 2x\,dx$. This perfectly matches the $2x\,dx$ portion of the original integral. Substituting gives us $\int \cos(u)\,du$, which is a basic integral that equals $\sin(u) + C$. Converting back to the original variable yields $\sin(x^2 + 1) + C$. You can verify this by differentiating: $\frac{d}{dx}[\sin(x^2 + 1)] = \cos(x^2 + 1) \cdot 2x = 2x\cos(x^2 + 1)$, which matches our original integrand. Answer choice A is incorrect because the substitution setup is actually perfect—$u = x^2 + 1$ and $du = 2x\,dx$ are both correct. Answer choice B is wrong since $\int \cos(u)\,du = \sin(u) + C$ is the correct antiderivative. Answer choice D is incorrect because neither step contains errors. The correct answer is C—the solution is completely correct throughout all steps. Study tip: When you see an integral containing both a function and what looks like its derivative, always check if substitution will work. Look for the "inside function" and its derivative appearing together.
6. Which of the following is the result of evaluating $\int (x^2 + 1)^2 dx$?
   Answer: $\frac{x^5}{5} + \frac{2x^3}{3} + x + C$ This integral should not be solved with u-substitution. Attempting to set $u = x^2+1$ yields $du = 2x \, dx$. Since there is no factor of $x$ in the integrand, substitution is not effective. Instead, the integrand should be expanded algebraically: $(x^2 + 1)^2 = (x^2)^2 + 2(x^2)(1) + 1^2 = x^4 + 2x^2 + 1$ Now, integrate the expanded polynomial term by term: $\int (x^4 + 2x^2 + 1) dx = \frac{x^5}{5} + \frac{2x^3}{3} + x + C$ Distractor A is a common error from misapplying the power rule directly to a composite function without accounting for the chain rule. Distractor B is an incorrect attempt to force a u-substitution. Distractor C results from forgetting the middle term when expanding the binomial square.
7. A company's revenue rate is $R'(t) = \frac{1200t}{(4t^2 + 9)^{3/2}}$ thousand dollars per month. Using u-substitution to find $R(t)$, if $u = 4t^2 + 9$, what is the coefficient that appears in front of the integral after substitution?
   Answer: $150$ When you encounter a u-substitution problem, you need to transform both the function and the differential to match your substitution. This question tests whether you can properly handle the coefficient that emerges during this transformation process. Given $u = 4t^2 + 9$, you first find $du = 8t \, dt$, which means $dt = \frac{du}{8t}$. Now you need to rewrite the original integral $\int \frac{1200t}{(4t^2 + 9)^{3/2}} dt$ in terms of $u$. Substituting $u = 4t^2 + 9$ and $dt = \frac{du}{8t}$: $\int \frac{1200t}{u^{3/2}} \cdot \frac{du}{8t} = \int \frac{1200t \cdot du}{u^{3/2} \cdot 8t} = \int \frac{1200}{8u^{3/2}} du = \int \frac{150}{u^{3/2}} du$ The coefficient in front of the integral after substitution is 150, making B correct. Looking at the wrong answers: A (300) likely comes from incorrectly calculating $\frac{1200}{8}$ or making an error with the differential substitution. C (600) might result from dividing 1200 by 2 instead of 8, perhaps confusing the derivative calculation. D (1200) suggests no recognition that substitution changes the coefficient—this would be the original coefficient before any transformation. Study tip: In u-substitution problems, always write out both $u =$ and $du =$ explicitly, then carefully substitute both parts. The coefficient almost always changes, so double-check your arithmetic when simplifying $\frac{\text{original coefficient}}{\text{coefficient from } du}$.
8. The integral $\int x^2(x^3 - 4)^7\,dx$ can be evaluated using u-substitution. What is the value of this integral?
   Answer: $\frac{1}{24}(x^3 - 4)^8 + C$ When you encounter an integral with a composition of functions like this, u-substitution is your go-to technique. The key is identifying the "inner function" and its derivative present in the integrand. Looking at $\int x^2(x^3 - 4)^7\,dx$, notice that we have $(x^3 - 4)$ raised to a power, and separately we have $x^2$. This suggests setting $u = x^3 - 4$, which means $\frac{du}{dx} = 3x^2$, or $du = 3x^2\,dx$. Since our integrand contains $x^2\,dx$, we can write $x^2\,dx = \frac{1}{3}du$. Substituting: $\int x^2(x^3 - 4)^7\,dx = \int u^7 \cdot \frac{1}{3}\,du = \frac{1}{3}\int u^7\,du$ Using the power rule: $\frac{1}{3} \cdot \frac{u^8}{8} = \frac{u^8}{24} = \frac{(x^3 - 4)^8}{24}$ Therefore, the answer is $\frac{1}{24}(x^3 - 4)^8 + C$, which is choice D. Choice A ignores the coefficient entirely and gets the wrong power. Choice B correctly identifies the power but misses the factor of 3 from the derivative. Choice C includes the factor of 3 in the numerator but forgets it should be in the denominator, suggesting someone multiplied by 3 instead of dividing. Study tip: In u-substitution problems, always check that the derivative of your chosen $u$ appears (possibly with a constant multiple) elsewhere in the integrand. Don't forget to account for any constant factors when converting $dx$ terms.
9. Determine the indefinite integral: $\int 4x^2 e^{x^3} dx$
   Answer: $\frac{4}{3}e^{x^3} + C$ We use u-substitution. Let $u = x^3$, the exponent of $e$. The derivative is $du = 3x^2 \, dx$. To match the integrand, we can solve for $x^2 \, dx$, which gives $x^2 \, dx = \frac{du}{3}$. Now substitute into the integral: $\int 4 e^u \left(\frac{du}{3}\right) = \frac{4}{3} \int e^u du$ The integral of $e^u$ is just $e^u$. So, we have: $\frac{4}{3} e^u + C$ Substituting back $u=x^3$ yields the final answer: $\frac{4}{3} e^{x^3} + C$ Distractor A results from forgetting the constant factor $\frac{1}{3}$ that arises from the substitution. Distractor B results from incorrectly multiplying by $3$ instead of dividing. Distractor D is the result of incorrectly integrating the $x^2$ term separately from the exponential term.
10. Which of the following integrals can be evaluated using a single u-substitution, and which substitution should be used?
    Answer: $\int x^2 e^{x^3}\,dx$ with $u = x^3$, but $\int x^2 e^{x^2}\,dx$ requires integration by parts When evaluating integrals involving exponentials and polynomials, the key is checking whether the derivative of the exponent appears as a factor in the integrand. This determines if u-substitution will work cleanly. For $\int x^2 e^{x^3}\,dx$, let's try $u = x^3$. Then $du = 3x^2\,dx$, so $x^2\,dx = \frac{1}{3}du$. The integral becomes $\frac{1}{3}\int e^u\,du = \frac{1}{3}e^u = \frac{1}{3}e^{x^3}$. This works perfectly because the $x^2$ factor is exactly what we need (up to a constant) for the substitution. For $\int x^2 e^{x^2}\,dx$, if we try $u = x^2$, then $du = 2x\,dx$. We have $x^2$ in our integrand, but we need just $x$ for the substitution. We can't cleanly separate out the extra factor of $x$, so simple u-substitution fails. This integral requires integration by parts. Choice A incorrectly claims the first integral works with $u = x^2$ - this substitution doesn't match the exponent $x^3$. Choice C wrongly states both integrals work with simple substitution. Choice D correctly identifies the substitution for the first integral but incorrectly suggests the second integral can use u-substitution followed by integration by parts. Study tip: For integrals like $\int f(x)e^{g(x)}\,dx$, u-substitution works when $f(x)$ is a constant multiple of $g'(x)$. If not, you'll likely need integration by parts or other advanced techniques.
11. If the substitution $u = 5x+2$ is used to evaluate the integral $\int \frac{3}{\sqrt{5x+2}} dx$, which of the following integrals is obtained?
    Answer: $\int \frac{3}{5\sqrt{u}} du$ When you encounter an integral that requires substitution, the key is remembering that you must substitute for both the variable AND the differential. The substitution affects every part of the integral. Given $u = 5x + 2$, you need to find what $dx$ becomes in terms of $du$. Taking the derivative: $\frac{du}{dx} = 5$, so $dx = \frac{1}{5}du$. Now substitute everything into the original integral $\int \frac{3}{\sqrt{5x+2}} dx$: - Replace $5x + 2$ with $u$ - Replace $dx$ with $\frac{1}{5}du$ This gives you: $\int \frac{3}{\sqrt{u}} \cdot \frac{1}{5} du = \int \frac{3}{5\sqrt{u}} du$ Looking at the wrong answers: Choice A forgot to account for the $dx$ substitution entirely—it kept the coefficient as 3 instead of adjusting for the $\frac{1}{5}$ factor. Choice B made the opposite error, multiplying by 5 instead of dividing by it ($3 \times 5 = 15$). Choice D unnecessarily complicated things by trying to express part of the integrand in terms of both $u$ and $x$, which defeats the purpose of substitution. Study tip: Always remember the substitution formula: if $u = g(x)$, then $du = g'(x)dx$, so $dx = \frac{du}{g'(x)}$. The derivative of your substitution function becomes the reciprocal of your adjustment factor. Practice identifying $g'(x)$ quickly—it's usually a constant in business calculus problems.
12. Evaluate the indefinite integral: $\int \frac{5x}{\sqrt{3x^2 + 7}} dx$
    Answer: $\frac{5}{3}\sqrt{3x^2 + 7} + C$ Let $u$ be the expression inside the square root: $u = 3x^2 + 7$. Then $du = 6x \, dx$. We can solve for $x \, dx$ to get $x \, dx = \frac{du}{6}$. Now substitute into the integral: $\int \frac{5}{\sqrt{u}} \left(\frac{du}{6}\right) = \frac{5}{6} \int u^{-1/2} du$ Apply the power rule for integration: $\frac{5}{6} \left( \frac{u^{1/2}}{1/2} \right) + C = \frac{5}{6} (2u^{1/2}) + C = \frac{5}{3}u^{1/2} + C$ Substitute back $u = 3x^2 + 7$: $\frac{5}{3}\sqrt{3x^2 + 7} + C$ Distractor B results from forgetting the $\frac{1}{6}$ from the $du$ term. Distractor C arises from an incorrect application of the power rule, subtracting 1 from the exponent instead of adding 1. Distractor D results from forgetting the factor of 5 from the original numerator.
13. Consider $\int \frac{\ln(x)}{x}\,dx$. If we use the substitution $u = \ln(x)$, what is the resulting integral in terms of $u$?
    Answer: $\int u\,du$ When you encounter an integral with a natural logarithm like this one, substitution is often the right approach. The key is systematically transforming both the integrand and the differential. Starting with $u = \ln(x)$, you need to find $du$. Taking the derivative: $\frac{du}{dx} = \frac{1}{x}$, so $du = \frac{1}{x}dx$. This is crucial because $\frac{1}{x}dx$ appears in your original integral. Now substitute into $\int \frac{\ln(x)}{x}dx$. Replace $\ln(x)$ with $u$ and $\frac{1}{x}dx$ with $du$: $\int \frac{u}{x} \cdot \frac{1}{x}dx = \int u \cdot du = \int u\,du$. The answer is D. Let's see why the other options miss the mark. Choice A ($\int \frac{1}{u}du$) incorrectly assumes the $\ln(x)$ in the numerator disappears, but our substitution replaces it with $u$. Choice B ($\int \frac{u}{x}du$) makes a common error—it partially substitutes but leaves $x$ in the denominator instead of recognizing that $\frac{1}{x}dx = du$. Choice C ($\int u \cdot e^u du$) incorrectly introduces an exponential function that doesn't exist in the original problem. Remember this pattern: when you see $\frac{\ln(x)}{x}$, the substitution $u = \ln(x)$ works perfectly because $du = \frac{1}{x}dx$ matches the remaining factor. Always ensure both the function and differential are completely transformed in your substitution.
14. To evaluate the integral $\int \frac{(\ln x)^3}{x} dx$, what is the most appropriate choice for $u$ and the resulting $du$?
    Answer: $u = \ln x$, $du = \frac{1}{x} dx$ The goal of u-substitution is to choose a part of the integrand, $u$, such that its derivative, $du$, also appears in the integrand (up to a constant factor). In the integral $\int \frac{(\ln x)^3}{x} dx = \int (\ln x)^3 \cdot \frac{1}{x} dx$, we can see two main functional parts: $(\ln x)^3$ and $\frac{1}{x}$. If we choose $u = \ln x$, its derivative is $du = \frac{1}{x} dx$. This perfectly matches the remaining part of the integrand. The integral then simplifies to $\int u^3 du$, which is easily solvable. Choice A is a possible substitution but does not simplify the integral effectively. Choice C is too complex; the resulting $du$ includes $(\ln x)^2$, which is not present in the original integrand to be substituted. Choice D is a trivial substitution that does not simplify the problem.
15. If $\int f'(g(x)) \cdot g'(x)\,dx = F(g(x)) + C$ where $F'(u) = f'(u)$, then $\int 3x^2 e^{x^3 + 2}\,dx$ equals:
    Answer: $e^{x^3 + 2} + C$ This integral has the form $\int f'(g(x)) \cdot g'(x)\,dx$ where $g(x) = x^3 + 2$ and $g'(x) = 3x^2$, and $f'(u) = e^u$ so $F(u) = e^u$. Using u-substitution with $u = x^3 + 2$, we get $du = 3x^2\,dx$. The integral becomes $\int e^u\,du = e^u + C = e^{x^3 + 2} + C$. Choice B incorrectly includes the coefficient 3 from the derivative. Choice C incorrectly applies $\frac{1}{3}$ as if compensating for a missing factor. Choice D incorrectly separates the exponent.
16. Find the indefinite integral: $\int x(x+3)^4 dx$
    Answer: $\frac{1}{6}(x+3)^6 - \frac{3}{5}(x+3)^5 + C$ This integral requires u-substitution where the remaining variable must also be expressed in terms of $u$. Let $u = x+3$. This implies that $du = dx$ and also that $x = u-3$. Substituting these into the integral gives: $\int (u-3)u^4 du$ Distribute $u^4$: $\int (u^5 - 3u^4) du$ Integrate term-by-term using the power rule: $\frac{u^6}{6} - \frac{3u^5}{5} + C$ Finally, substitute back $u=x+3$: $\frac{1}{6}(x+3)^6 - \frac{3}{5}(x+3)^5 + C$ Distractor B results from a sign error, using $x=3-u$ instead of $x=u-3$. Distractor C incorrectly integrates the parts of the product separately. Distractor D results from an error in distributing $u^4$ across $(u-3)$.
17. Find the indefinite integral: $\int \frac{8x - 4}{x^2 - x + 5} dx$
    Answer: $4 \ln|x^2 - x + 5| + C$ This integral is in a form that leads to a natural logarithm. Let $u$ be the denominator, $u = x^2 - x + 5$. Then the differential is $du = (2x - 1) \, dx$. We can factor the numerator of the integrand to relate it to $du$: $8x - 4 = 4(2x - 1)$. Now, we can rewrite the integral and substitute: $\int \frac{4(2x - 1)}{x^2 - x + 5} dx = \int \frac{4}{u} du$ Integrate with respect to $u$: $4 \int \frac{1}{u} du = 4 \ln|u| + C$ Finally, substitute back $u=x^2-x+5$: $4 \ln|x^2 - x + 5| + C$ Distractor B results from correctly identifying the substitution but forgetting the constant factor of 4. Distractor C arises from incorrectly applying the power rule to $(x^2 - x + 5)^{-1}$. Distractor D is the result of improperly integrating the numerator and denominator separately.
18. Evaluate the definite integral: $\int_{0}^{a} x(a^2 - x^2)^3 dx$, where $a$ is a positive constant.
    Answer: $\frac{a^8}{8}$ When you encounter a definite integral with a polynomial expression raised to a power, look for substitution opportunities that can simplify the problem dramatically. For $\int_{0}^{a} x(a^2 - x^2)^3 dx$, notice that the derivative of $(a^2 - x^2)$ is $-2x$, which relates directly to the $x$ factor outside the parentheses. This suggests using $u$-substitution with $u = a^2 - x^2$. Setting up the substitution: $u = a^2 - x^2$, so $du = -2x \, dx$, which means $x \, dx = -\frac{1}{2}du$. For the limits: when $x = 0$, $u = a^2 - 0^2 = a^2$; when $x = a$, $u = a^2 - a^2 = 0$. The integral becomes: $\int_{a^2}^{0} u^3 \left(-\frac{1}{2}\right) du = \frac{1}{2}\int_{0}^{a^2} u^3 du$ Evaluating: $\frac{1}{2} \left[\frac{u^4}{4}\right]_0^{a^2} = \frac{1}{2} \cdot \frac{(a^2)^4}{4} = \frac{1}{2} \cdot \frac{a^8}{4} = \frac{a^8}{8}$ This confirms answer B is correct. Answer A ($-\frac{a^8}{8}$) results from forgetting to flip the limits when substituting or making a sign error. Answer C ($\frac{a^8}{4}$) comes from missing the factor of $\frac{1}{2}$ in the substitution. Answer D ($\frac{a^6}{6}$) suggests incorrectly integrating the original expression without proper substitution. Strategy tip: When you see $x$ multiplied by an expression containing $x^2$, immediately check if $u$-substitution can eliminate the $x$ factor—it often transforms complex-looking integrals into simple polynomial integration.
19. Find the indefinite integral: $\int x^3 \sqrt{x^2+4} \, dx$
    Answer: $\frac{1}{5}(x^2+4)^{5/2} - \frac{4}{3}(x^2+4)^{3/2} + C$ When you encounter an integral with a polynomial multiplied by a square root expression, substitution is typically your best approach. Here, the presence of $x^3 \sqrt{x^2+4}$ suggests using $u = x^2 + 4$. Let $u = x^2 + 4$, so $du = 2x \, dx$ or $x \, dx = \frac{1}{2}du$. Since we have $x^3$, we can rewrite it as $x^2 \cdot x = (u-4) \cdot x$. The integral becomes: $\int x^3 \sqrt{x^2+4} \, dx = \int x^2 \cdot x \sqrt{x^2+4} \, dx = \int (u-4) \sqrt{u} \cdot \frac{1}{2} du$ $= \frac{1}{2} \int (u-4) u^{1/2} \, du = \frac{1}{2} \int (u^{3/2} - 4u^{1/2}) \, du$ $= \frac{1}{2}\left[\frac{u^{5/2}}{5/2} - 4 \cdot \frac{u^{3/2}}{3/2}\right] + C = \frac{1}{5}u^{5/2} - \frac{4}{3}u^{3/2} + C$ Substituting back: $\frac{1}{5}(x^2+4)^{5/2} - \frac{4}{3}(x^2+4)^{3/2} + C$ Answer A missing the negative term shows incomplete integration—likely forgetting to distribute the $(u-4)$ properly. Answer B has the correct structure but wrong coefficients, possibly from arithmetic errors in the substitution. Answer D attempts integration by parts incorrectly and doesn't complete the process. Study tip: For integrals involving $x^n \sqrt{ax^2 + b}$, always try $u = ax^2 + b$ first. Remember to express all powers of $x$ in terms of your substitution variable before integrating.
20. Evaluate the definite integral: $\int_{1}^{2} \frac{6x}{(x^2+2)^3} dx$
    Answer: $\frac{1}{8}$ To evaluate the integral, we use u-substitution. Let $u = x^2+2$. Then, the derivative is $du = 2x \, dx$. The term $6x \, dx$ in the integrand can be written as $3 \cdot (2x \, dx) = 3 \, du$. We must also change the limits of integration from $x$-values to $u$-values. The lower bound is $x=1$, so $u = 1^2+2 = 3$. The upper bound is $x=2$, so $u = 2^2+2 = 6$. The integral transforms to: $\int_{3}^{6} \frac{3}{u^3} du = \int_{3}^{6} 3u^{-3} du$ Integrating with respect to $u$ gives: $3 \left[ \frac{u^{-2}}{-2} \right]_{3}^{6} = -\frac{3}{2} \left[ \frac{1}{u^2} \right]_{3}^{6}$ Now, we evaluate at the new bounds: $-\frac{3}{2} \left( \frac{1}{6^2} - \frac{1}{3^2} \right) = -\frac{3}{2} \left( \frac{1}{36} - \frac{1}{9} \right) = -\frac{3}{2} \left( \frac{1}{36} - \frac{4}{36} \right) = -\frac{3}{2} \left( -\frac{3}{36} \right) = -\frac{3}{2} \left( -\frac{1}{12} \right) = \frac{3}{24} = \frac{1}{8}$

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