Integration: Long Division & Completing Square - Calculus 1

1. For what values of the constant $b$ does the integral $\int \frac{1}{x^2+bx+1} dx$ result in an expression involving an arctangent function, rather than a natural logarithm?
   Answer: $-2 < b < 2$ When you encounter an integral of the form $\int \frac{1}{x^2+bx+1} dx$, the key is recognizing that the result depends on whether the quadratic denominator can be factored into real linear terms or not. This determines whether you'll use partial fractions (leading to logarithms) or complete the square (leading to arctangent). The quadratic $x^2+bx+1$ factors into real linear terms when its discriminant is positive: $b^2-4(1)(1) > 0$, or $b^2 > 4$. When this happens, you can use partial fractions, resulting in natural logarithms. However, when the discriminant is negative ($b^2 < 4$), the quadratic has no real roots, so you complete the square and get an arctangent form. For an arctangent result, you need $b^2 < 4$, which means $-2 < b < 2$. This makes choice A correct. Choice B ($b = ±2$) represents the boundary cases where $b^2 = 4$, giving a perfect square denominator and resulting in a simple rational function, not arctangent. Choice C incorrectly suggests the form doesn't depend on $b$'s value. Choice D ($b > 2$ or $b < -2$) gives the conditions for logarithmic results, which is the opposite of what we want. Study tip: Remember the discriminant rule for rational function integration: negative discriminant → arctangent, positive discriminant → logarithm. The boundary between these regions (where the discriminant equals zero) often appears as a distractor choice.
2. To evaluate $\int \frac{dx}{ax^2+bx+c}$ where $b^2-4ac < 0$ and $a>0$, completing the square transforms the denominator into the form $a[(x+k)^2+h^2]$. What is $h$ in terms of $a, b, c$?
   Answer: $\frac{\sqrt{4ac-b^2}}{2a}$ When you encounter integrals with quadratic denominators where the discriminant is negative, completing the square is essential to transform the expression into a form that leads to an inverse tangent antiderivative. To complete the square for $ax^2 + bx + c$, you factor out the coefficient of $x^2$ first: $a(x^2 + \frac{b}{a}x + \frac{c}{a})$. Then complete the square inside the parentheses by adding and subtracting $(\frac{b}{2a})^2$: $a\left[x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} + \frac{c}{a}\right]$ This becomes: $a\left[\left(x + \frac{b}{2a}\right)^2 + \frac{c}{a} - \frac{b^2}{4a^2}\right]$ Simplifying the constant term: $\frac{c}{a} - \frac{b^2}{4a^2} = \frac{4ac - b^2}{4a^2}$ So we have: $a\left[\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a^2}\right]$ Comparing with $a[(x+k)^2 + h^2]$, we see that $h^2 = \frac{4ac - b^2}{4a^2}$, so $h = \frac{\sqrt{4ac - b^2}}{2a}$, which is choice B. Choice A uses $\sqrt{b^2 - 4ac}$, which would be imaginary since $b^2 - 4ac < 0$. Choice C gives you $k$, not $h$. Choice D has the wrong denominator—it should be $2a$, not $a$. Remember: when $b^2 - 4ac < 0$, always flip the sign under the square root to $4ac - b^2$ during completing the square to ensure you're working with real numbers.
3. Which expression represents the integral $\int \frac{1}{2x^2 + 4x + 10} dx$?
   Answer: $\frac{1}{4}\arctan\left(\frac{x+1}{2}\right) + C$ First, factor out the leading coefficient from the denominator: $\int \frac{1}{2(x^2 + 2x + 5)} dx = \frac{1}{2} \int \frac{1}{x^2 + 2x + 5} dx$. Next, complete the square for the quadratic in the denominator: $x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x+1)^2 + 2^2$. The integral becomes $\frac{1}{2} \int \frac{1}{(x+1)^2 + 2^2} dx$. This is an arctangent form $\frac{1}{a}\arctan(\frac{u}{a})$ with $u=x+1$ and $a=2$. Integrating gives: $\frac{1}{2} \left[ \frac{1}{2} \arctan\left(\frac{x+1}{2}\right) \right] + C = \frac{1}{4} \arctan\left(\frac{x+1}{2}\right) + C$.
4. Which of the following is equivalent to $\int \frac{2x - 1}{x^2 - 6x + 13} dx$?
   Answer: $\ln(x^2 - 6x + 13) + \frac{5}{2}\arctan\left(\frac{x-3}{2}\right) + C$ The denominator $x^2 - 6x + 13$ is an irreducible quadratic. First, complete the square: $x^2 - 6x + 13 = (x^2 - 6x + 9) + 4 = (x-3)^2 + 2^2$. The derivative of the original denominator is $2x-6$. We manipulate the numerator to contain this term: $2x - 1 = (2x - 6) + 5$. Now, split the integral: $\int \frac{(2x - 6) + 5}{x^2 - 6x + 13} dx = \int \frac{2x - 6}{x^2 - 6x + 13} dx + \int \frac{5}{(x-3)^2 + 4} dx.$ The first integral is solved with a u-substitution $u = x^2 - 6x + 13$, yielding $\ln|x^2 - 6x + 13|$. The quadratic is always positive, so the absolute value is not needed. The second integral is an arctangent form: $5 \cdot \frac{1}{2}\arctan\left(\frac{x-3}{2}\right)$. Combining these gives $\ln(x^2 - 6x + 13) + \frac{5}{2}\arctan\left(\frac{x-3}{2}\right) + C$.
5. The expression $\frac{x^4+3x^2+1}{x^2+1}$ can be simplified to $P(x) + \frac{C}{x^2+1}$ where $P(x)$ is a polynomial and $C$ is a constant. Find $\int_0^{\sqrt{3}} \frac{x^4+3x^2+1}{x^2+1} dx$.
   Answer: $3\sqrt{3} - \frac{\pi}{3}$ When you encounter a rational function where the degree of the numerator is greater than or equal to the degree of the denominator, you need to perform polynomial long division first. This separates the function into a polynomial part plus a simpler rational remainder. Let's divide $x^4 + 3x^2 + 1$ by $x^2 + 1$. Since there's no $x^3$ term in the numerator, we get: - $x^4 ÷ x^2 = x^2$, giving $x^2(x^2 + 1) = x^4 + x^2$ - Subtracting: $(x^4 + 3x^2 + 1) - (x^4 + x^2) = 2x^2 + 1$ - $2x^2 ÷ x^2 = 2$, giving $2(x^2 + 1) = 2x^2 + 2$ - Subtracting: $(2x^2 + 1) - (2x^2 + 2) = -1$ So $\frac{x^4+3x^2+1}{x^2+1} = x^2 + 2 + \frac{-1}{x^2+1}$ Now integrate: $\int_0^{\sqrt{3}} \left(x^2 + 2 - \frac{1}{x^2+1}\right) dx$ This becomes: $\left[\frac{x^3}{3} + 2x - \arctan(x)\right]_0^{\sqrt{3}}$ Evaluating: $\left(\frac{(\sqrt{3})^3}{3} + 2\sqrt{3} - \arctan(\sqrt{3})\right) - (0 + 0 - 0)$ $= \frac{3\sqrt{3}}{3} + 2\sqrt{3} - \frac{\pi}{3} = \sqrt{3} + 2\sqrt{3} - \frac{\pi}{3} = 3\sqrt{3} - \frac{\pi}{3}$ Answer B is correct. Answer A adds instead of subtracts the arctangent term. Answer C miscalculates the polynomial integration. Answer D uses $\frac{\pi}{6}$ instead of $\frac{\pi}{3}$ for $\arctan(\sqrt{3})$. Study tip: Always perform polynomial long division when the numerator's degree isn't smaller than the denominator's, and remember that $\arctan(\sqrt{3}) = \frac{\pi}{3}$.
6. The integral $\int \frac{x^n}{x-1} dx$ for an integer $n \geq 1$ results in a function of the form $P(x) + \ln|x-1| + C$, where $P(x)$ is a polynomial. What is the degree of $P(x)$?
   Answer: $n$ To integrate $\frac{x^n}{x-1}$, we perform polynomial long division. The division of $x^n$ by $x-1$ results in a quotient and a remainder. Using the remainder theorem, the remainder is $(1)^n=1$. The quotient, $Q(x)$, will have a degree of $n-1$. \ $\frac{x^n}{x-1} = Q(x) + \frac{1}{x-1}$ \ where $Q(x) = x^{n-1} + x^{n-2} + \dots + x + 1$. \ The integral is then: \ $\int \left(x^{n-1} + x^{n-2} + \dots + x + 1 + \frac{1}{x-1}\right) dx$ \ The term $P(x)$ in the result corresponds to the integral of the polynomial $Q(x)$: \ $P(x) = \int Q(x) dx = \int (x^{n-1} + x^{n-2} + \dots + 1) dx$ \ $P(x) = \frac{x^n}{n} + \frac{x^{n-1}}{n-1} + \dots + x$ \ The highest power of $x$ in $P(x)$ is $n$, so the degree of the polynomial $P(x)$ is $n$.
7. A certain calculation requires finding the antiderivative of $f(x) = \frac{x+1}{4x^2+1}$. Another calculation requires finding the antiderivative of $g(x) = \frac{x^3}{4x^2+1}$. Which of the following statements is true?
   Answer: Evaluating $\int g(x) dx$ requires long division, while $\int f(x) dx$ does not. When you encounter rational functions where the degree of the numerator is greater than or equal to the degree of the denominator, you need to consider whether polynomial long division is necessary before integration. For $\int f(x) dx = \int \frac{x+1}{4x^2+1} dx$, the numerator has degree 1 and denominator has degree 2, so the degree of the numerator is less than the denominator. This means you can integrate directly using substitution and partial fraction techniques. You can split this as $\int \frac{x}{4x^2+1} dx + \int \frac{1}{4x^2+1} dx$. The first integral uses u-substitution with $u = 4x^2+1$, and the second is an arctangent form. For $\int g(x) dx = \int \frac{x^3}{4x^2+1} dx$, the numerator has degree 3 and denominator has degree 2. Since the numerator's degree is higher, you must perform polynomial long division first. Dividing $x^3$ by $4x^2+1$ gives you $\frac{1}{4}x - \frac{1}{16} + \frac{\frac{1}{16}}{4x^2+1}$, which you can then integrate term by term. Option A incorrectly states that $f(x)$ requires long division when it doesn't. Option C is wrong because only $g(x)$ needs long division. Option D is incorrect because $g(x)$ cannot be solved by direct substitution alone due to the degree issue. Study tip: Always compare the degrees of numerator and denominator first. If the numerator's degree is greater than or equal to the denominator's degree, perform long division before attempting integration.
8. Which of the following is equivalent to $\int \frac{2x-3}{x^2+6x+10} dx$?
   Answer: $\ln(x^2+6x+10) - 9\arctan(x+3) + C$ The integrand's denominator is an irreducible quadratic. The derivative of $x^2+6x+10$ is $2x+6$. We manipulate the numerator to contain this expression: \ $2x-3 = (2x+6) - 9$ \ This allows us to split the integral into two parts: \ $\int \frac{2x+6}{x^2+6x+10} dx - \int \frac{9}{x^2+6x+10} dx$ \ The first integral is a standard natural log form, yielding $\ln(x^2+6x+10)$ (absolute value is not needed as the quadratic is always positive). For the second integral, we complete the square in the denominator: \ $x^2+6x+10 = (x^2+6x+9)+1 = (x+3)^2+1$ \ So the second integral is $-9 \int \frac{1}{(x+3)^2+1} dx = -9\arctan(x+3)$. \ Combining the parts gives $\ln(x^2+6x+10) - 9\arctan(x+3) + C$.
9. Let $g(x)$ be the antiderivative of $\frac{4x^3+6x^2+1}{2x^2+1}$. Which of the following could be $g(x)$?
   Answer: $x^2+3x - \frac{1}{2}\ln(2x^2+1) - \sqrt{2}\arctan(\sqrt{2}x) + C$ When you encounter a rational function to integrate, you'll often need to use polynomial long division followed by partial fraction decomposition or substitution techniques. To find the antiderivative of $\frac{4x^3+6x^2+1}{2x^2+1}$, first perform polynomial long division since the numerator's degree is higher than the denominator's. Dividing $4x^3+6x^2+1$ by $2x^2+1$: $\frac{4x^3+6x^2+1}{2x^2+1} = 2x + 3 + \frac{-4x-2}{2x^2+1}$ Now you need to integrate each part: $\int(2x + 3)dx + \int\frac{-4x-2}{2x^2+1}dx$ The first integral gives $x^2 + 3x$. For the second, split it: $\int\frac{-4x}{2x^2+1}dx + \int\frac{-2}{2x^2+1}dx$ The first uses substitution ($u = 2x^2+1$) yielding $-\ln(2x^2+1)$. The second requires the arctangent formula, giving $-\sqrt{2}\arctan(\sqrt{2}x)$. Combining everything: $g(x) = x^2 + 3x - \ln(2x^2+1) - \sqrt{2}\arctan(\sqrt{2}x) + C$ But wait—this matches option B, not C! The key difference is the coefficient of the logarithm term. When integrating $\frac{-4x}{2x^2+1}$, you get $-\frac{1}{2}\ln(2x^2+1)$, not $-\ln(2x^2+1)$. Option A has the wrong polynomial part. Option B has an incorrect logarithm coefficient. Option D has wrong signs throughout. Study tip: Always double-check your polynomial long division and carefully track coefficients when using substitution—these are the most common sources of error in rational function integration.
10. Let $f(x) = \frac{1}{x^2-4x+5}$ and $g(x) = \frac{1}{\sqrt{2x-x^2}}$. Let $A = \int_2^3 f(x)dx$ and $B = \int_1^{1.5} g(x)dx$. Which of the following describes the relationship between A and B?
    Answer: $A = \frac{\pi}{4}, B = \frac{\pi}{6}$ When you encounter integrals with expressions like $\frac{1}{x^2-4x+5}$ or $\frac{1}{\sqrt{2x-x^2}}$, look for opportunities to complete the square and use inverse trigonometric function formulas. For integral A, complete the square in the denominator: $x^2-4x+5 = (x-2)^2+1$. So $A = \int_2^3 \frac{1}{(x-2)^2+1}dx$. Using substitution $u = x-2$, when $x=2$, $u=0$; when $x=3$, $u=1$. This gives us $A = \int_0^1 \frac{1}{u^2+1}du = \arctan(u)\Big|_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4}$. For integral B, complete the square under the square root: $2x-x^2 = -(x^2-2x) = -(x-1)^2+1$. So $B = \int_1^{1.5} \frac{1}{\sqrt{1-(x-1)^2}}dx$. Using substitution $u = x-1$, when $x=1$, $u=0$; when $x=1.5$, $u=0.5$. This gives us $B = \int_0^{0.5} \frac{1}{\sqrt{1-u^2}}du = \arcsin(u)\Big|_0^{0.5} = \arcsin(0.5) - \arcsin(0) = \frac{\pi}{6}$. Therefore $A = \frac{\pi}{4}$ and $B = \frac{\pi}{6}$, making B correct. Choice A reverses these values. Choice C stops at the intermediate step before evaluating the special angles. Choice D incorrectly suggests these are logarithmic integrals rather than inverse trigonometric ones. Remember: expressions of the form $\frac{1}{u^2+1}$ integrate to $\arctan(u)$, while $\frac{1}{\sqrt{1-u^2}}$ integrates to $\arcsin(u)$. Complete the square first to recognize these patterns.
11. Evaluate the definite integral $\int_0^1 \frac{x^2}{x+1} dx$.
    Answer: $\ln(2) - \frac{1}{2}$ When you encounter a rational function where the degree of the numerator is close to the degree of the denominator, polynomial long division is often your best approach. Here, we have $\frac{x^2}{x+1}$, where both numerator and denominator are degree 1 apart. Dividing $x^2$ by $(x+1)$: - $x^2 = (x+1) \cdot x - x$ - $-x = (x+1) \cdot (-1) + 1$ So $\frac{x^2}{x+1} = x - 1 + \frac{1}{x+1}$ Now we can integrate term by term: $\int_0^1 \frac{x^2}{x+1} dx = \int_0^1 \left(x - 1 + \frac{1}{x+1}\right) dx$ $= \left[\frac{x^2}{2} - x + \ln|x+1|\right]_0^1$ $= \left(\frac{1}{2} - 1 + \ln(2)\right) - \left(0 - 0 + \ln(1)\right)$ $= -\frac{1}{2} + \ln(2) = \ln(2) - \frac{1}{2}$ Answer D is correct. Answer A gives $-\frac{1}{2}$, which would result from forgetting the logarithmic term entirely. Answer B gives $\ln(2) + \frac{1}{2}$, a sign error when computing $\frac{1}{2} - 1$. Answer C gives $1 - \ln(2)$, likely from incorrectly handling the polynomial division or making multiple arithmetic errors. Study tip: When integrating rational functions, always check if polynomial long division simplifies the problem first. This technique transforms difficult integrals into manageable polynomial plus simple rational terms.
12. For what value of $b > 0$ does $\int_0^b \frac{2x}{x^2+1} dx = \int_0^1 \frac{2x^3+2x+1}{x^2+1} dx$?
    Answer: $\sqrt{e^{1+\pi/4} - 1}$ When you encounter equations with definite integrals, you need to evaluate each integral separately and then solve for the unknown parameter. This question tests your ability to compute integrals and work with logarithmic equations. First, let's evaluate the left integral. For $\int_0^b \frac{2x}{x^2+1} dx$, notice that the derivative of $x^2+1$ is $2x$, so this is a perfect setup for substitution. Let $u = x^2+1$, then $du = 2x\,dx$. This gives us $\int \frac{1}{u} du = \ln|u| = \ln(x^2+1)$. Evaluating from 0 to $b$: $\ln(b^2+1) - \ln(1) = \ln(b^2+1)$. For the right integral, split $\frac{2x^3+2x+1}{x^2+1}$ by polynomial long division: $\frac{2x^3+2x+1}{x^2+1} = 2x + \frac{1}{x^2+1}$. So $\int_0^1 \left(2x + \frac{1}{x^2+1}\right) dx = [x^2 + \arctan(x)]_0^1 = 1 + \frac{\pi}{4}$. Setting the integrals equal: $\ln(b^2+1) = 1 + \frac{\pi}{4}$. Solving for $b$: $b^2+1 = e^{1+\pi/4}$, so $b^2 = e^{1+\pi/4} - 1$, giving $b = \sqrt{e^{1+\pi/4} - 1}$. Choice A omits the square root, choice B uses only $\pi/4$ in the exponent (missing the 1), and choice C uses only $e$ (missing the $\pi/4$). Study tip: When comparing definite integrals, evaluate each separately first, then solve the resulting equation. Watch for substitution opportunities when the numerator is the derivative of the denominator.
13. Evaluate $\int_1^2 \frac{2x+2}{x^2-2x+2} dx$.
    Answer: $\pi + \ln(2)$ First, complete the square in the denominator: $x^2-2x+2 = (x-1)^2+1$. The derivative of the denominator is $2x-2$. Rewrite the numerator as $2x+2 = (2x-2)+4$. Split the integral: $\int_1^2 \frac{2x-2}{x^2-2x+2} dx + \int_1^2 \frac{4}{(x-1)^2+1} dx.$ The first integral evaluates to $[\ln|x^2-2x+2|]_1^2 = \ln(2) - \ln(1) = \ln(2)$. The second integral evaluates to $[4\arctan(x-1)]_1^2 = 4\arctan(1) - 4\arctan(0) = 4(\frac{\pi}{4}) - 0 = \pi$. The sum of the two parts is $\pi + \ln(2)$.
14. Evaluate the integral $\int \frac{1}{\sqrt{-x^2 + 8x - 12}} dx$.
    Answer: $\arcsin\left(\frac{x-4}{2}\right) + C$ The integral suggests an inverse sine function, which requires completing the square in the denominator. First, factor out -1 from the quadratic: $-x^2 + 8x - 12 = -(x^2 - 8x + 12)$. Complete the square for $x^2 - 8x$: $x^2 - 8x + 16 = (x-4)^2$. So, $-(x^2 - 8x + 16 - 16 + 12) = -((x-4)^2 - 4) = 4 - (x-4)^2$. The integral becomes $\int \frac{1}{\sqrt{4 - (x-4)^2}} dx = \int \frac{1}{\sqrt{2^2 - (x-4)^2}} dx$. This is the standard form for arcsin, $\int \frac{1}{\sqrt{a^2 - u^2}} du = \arcsin(\frac{u}{a}) + C$, with $a=2$ and $u=x-4$. The result is $\arcsin\left(\frac{x-4}{2}\right) + C$.
15. The evaluation of $\int \frac{x^3}{x^2 - 2x + 2} dx$ requires multiple steps. Which of the following correctly represents the integral after performing the initial polynomial division?
    Answer: $\int \left(x + 2 + \frac{2x - 4}{x^2 - 2x + 2}\right) dx$ Since the degree of the numerator (3) is greater than the degree of the denominator (2), the first step is polynomial long division. Dividing $x^3$ by $x^2 - 2x + 2$: $$\begin{array}{r} x + 2 \\ x^2 - 2x + 2 \overline{) x^3 + 0x^2 + 0x + 0} \\ - (x^3 - 2x^2 + 2x) \\ \hline 2x^2 - 2x + 0 \\ - (2x^2 - 4x + 4) \\ \hline 2x - 4 \end{array}$$ The quotient is $x+2$ and the remainder is $2x-4$. Therefore, the original integrand can be rewritten as $x + 2 + \frac{2x - 4}{x^2 - 2x + 2}$. The integral becomes $\int \left(x + 2 + \frac{2x - 4}{x^2 - 2x + 2}\right) dx$.
16. Find the area of the region bounded by the graph of $f(x) = \frac{1}{x^2 - 4x + 13}$, the x-axis, and the lines $x=2$ and $x=5$.
    Answer: $\frac{\pi}{12}$ The area is given by the definite integral $A = \int_{2}^{5} \frac{1}{x^2 - 4x + 13} dx$. The denominator is an irreducible quadratic, so we complete the square: \ $x^2 - 4x + 13 = (x^2 - 4x + 4) + 9 = (x-2)^2 + 3^2$ \ The integral becomes $\int_{2}^{5} \frac{1}{(x-2)^2 + 3^2} dx$. Let $u = x-2$, so $du = dx$. The limits of integration change from $x=2, 5$ to $u=0, 3$. \ $A = \int_{0}^{3} \frac{1}{u^2 + 3^2} du = \left[ \frac{1}{3}\arctan\left(\frac{u}{3}\right) \right]_{0}^{3}$ \ Evaluating at the limits gives: \ $\frac{1}{3}\arctan\left(\frac{3}{3}\right) - \frac{1}{3}\arctan\left(\frac{0}{3}\right) = \frac{1}{3}\arctan(1) - 0 = \frac{1}{3} \cdot \frac{\pi}{4} = \frac{\pi}{12}$
17. Consider the integral $I = \int \frac{P(x)}{Q(x)} dx$, where $P(x)$ is a polynomial of degree 4, $Q(x)$ is a polynomial of degree 2, and $Q(x)$ is irreducible. Which sequence of steps is the most direct and appropriate method to evaluate $I$?
    Answer: Perform long division, then complete the square on the denominator of the resulting rational term and split this term into logarithmic and arctangent parts for integration. Since the degree of the numerator $P(x)$ (4) is greater than or equal to the degree of the denominator $Q(x)$ (2), the first required step is polynomial long division. This will result in an expression of the form $\text{Quotient} + \frac{\text{Remainder}}{Q(x)}$, where the degree of the Remainder is less than 2. The Quotient is a polynomial and can be integrated easily. The term $\frac{\text{Remainder}}{Q(x)}$ has an irreducible quadratic denominator, so the next step is to complete the square in $Q(x)$. After completing the square, the fraction is typically split into two parts: one that integrates to a natural logarithm and one that integrates to an arctangent.
18. Which integral is generated by an appropriate u-substitution after preparing $\int \frac{dx}{\sqrt{3-4x-4x^2}}$ for integration?
    Answer: $\frac{1}{2} \int \frac{1}{\sqrt{4-u^2}} du$ When you encounter an integral with a quadratic expression under a square root, you need to complete the square first, then use u-substitution to transform it into a standard form. Starting with $\int \frac{dx}{\sqrt{3-4x-4x^2}}$, first factor out the coefficient of $x^2$ from the quadratic: $3-4x-4x^2 = -4(x^2+x) + 3$. To complete the square for $x^2+x$, add and subtract $(\frac{1}{2})^2 = \frac{1}{4}$: $x^2+x = (x+\frac{1}{2})^2 - \frac{1}{4}$. Substituting back: $3-4x-4x^2 = -4[(x+\frac{1}{2})^2 - \frac{1}{4}] + 3 = -4(x+\frac{1}{2})^2 + 1 + 3 = 4 - 4(x+\frac{1}{2})^2$. The integral becomes $\int \frac{dx}{\sqrt{4 - 4(x+\frac{1}{2})^2}}$. Now substitute $u = 2(x+\frac{1}{2})$, so $du = 2dx$ or $dx = \frac{1}{2}du$. Also, $4(x+\frac{1}{2})^2 = u^2$. This gives us $\int \frac{\frac{1}{2}du}{\sqrt{4-u^2}} = \frac{1}{2}\int \frac{du}{\sqrt{4-u^2}}$, which matches answer D. Answer A has the wrong sign ($u^2+4$ instead of $4-u^2$) and represents a different standard form. Answer B is missing the factor of $\frac{1}{2}$ that comes from the u-substitution. Answer C has both the wrong sign under the square root and represents yet another standard form entirely. Study tip: Always complete the square first for quadratic expressions under radicals, then carefully track coefficients during u-substitution to avoid losing constant factors.
19. Let $F(x) = \int \frac{x^4}{x^2+1} dx$. If $F(0) = 1$, what is the value of $F(1)$?
    Answer: $\frac{1}{3} + \frac{\pi}{4}$ First, perform long division on $\frac{x^4}{x^2+1}$. A quick way is to rewrite the numerator: $\frac{x^4}{x^2+1} = \frac{x^4-1+1}{x^2+1} = \frac{(x^2-1)(x^2+1)+1}{x^2+1} = x^2-1 + \frac{1}{x^2+1}$. Now, integrate: $F(x) = \int (x^2-1 + \frac{1}{x^2+1}) dx = \frac{x^3}{3} - x + \arctan(x) + C$. Use the initial condition $F(0) = 1$: $\frac{0}{3} - 0 + \arctan(0) + C = 1 \implies C=1$. So, $F(x) = \frac{x^3}{3} - x + \arctan(x) + 1$. Finally, evaluate at $x=1$: $F(1) = \frac{1^3}{3} - 1 + \arctan(1) + 1 = \frac{1}{3} + \frac{\pi}{4}$.
20. Which of the following describes a necessary step in evaluating $\int \frac{x-3}{x^2+x+1} dx$?
    Answer: Rewriting the numerator as a combination of the derivative of the denominator and a constant. When you encounter a rational function where the degree of the numerator is less than the degree of the denominator, and the denominator cannot be factored into linear terms, you need a strategic approach to integration. For $\int \frac{x-3}{x^2+x+1} dx$, the key insight is to rewrite the numerator in terms of the derivative of the denominator. Since the derivative of $x^2+x+1$ is $2x+1$, you can express $x-3$ as $\frac{1}{2}(2x+1) + c$ for some constant $c$. Working this out: $x-3 = \frac{1}{2}(2x+1) - \frac{7}{2}$. This splits the integral into $\frac{1}{2}\int \frac{2x+1}{x^2+x+1} dx - \frac{7}{2}\int \frac{1}{x^2+x+1} dx$, where the first part becomes a simple logarithm and the second requires completing the square. Option A is incorrect because $x^2+x+1$ cannot be factored into linear factors with real coefficients (its discriminant is negative), so partial fraction decomposition with linear denominators is impossible. Option B is wrong because the degree of the numerator (1) is already less than the degree of the denominator (2), so polynomial long division is unnecessary. Option D fails because substituting $u = x^2+x+1$ makes $du = (2x+1)dx$, but our numerator is $x-3$, not $2x+1$, so this substitution doesn't work directly. Remember: when integrating rational functions where the denominator can't be factored nicely, always try expressing the numerator in terms of the denominator's derivative plus a constant.

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