Integrating Using Substitution - AP Calculus AB

Compute $\int \dfrac{7}{\sqrt{7x+1}}\,dx$ for a quantity depending on $\sqrt{7x+1}$ .
Answer: $2\sqrt{7x+1}+C$ This integral uses u-substitution with a square root in the denominator. Rewriting as $7(7x+1)^{-1/2}$ , we see the inner function is $7x + 1$ with derivative 7. Setting $u = 7x + 1$ gives $du = 7\,dx$ , so the integral becomes $\int u^{-1/2}\,du = 2u^{1/2} + C$ . Substituting back yields $2\sqrt{7x+1} + C$ . Choice A incorrectly multiplies by an extra factor of $\frac{7}{2}$ , confusing the coefficient in the substitution with the power rule result. When integrating functions of the form $\frac{a}{\sqrt{ax+b}}$ , the result is $2\sqrt{ax+b} + C$ .
Evaluate $\int \dfrac{\cos(5x)}{1+\sin(5x)}\,5\,dx$ .
Answer: $\ln|1+\sin(5x)|+C$ The skill here is integrating using u-substitution. The denominator is 1 + sin(5x), with the numerator effectively 5 cos(5x) from the chain rule. Set u = 1 + sin(5x) with du = 5 cos(5x) dx, giving ∫ du/u = ln|u| + C. This is ln|1 + sin(5x)| + C. Choice B divides by 5 extra, mishandling the du coefficient. Recognize trigonometric compositions with chain rule derivatives in numerators.
Compute $\int \dfrac{3x^2+2x}{x^3+x^2+4}\,dx$ for a logarithmic accumulation term.
Answer: $\ln|x^3+x^2+4|+C$ This integral requires the skill of integration by substitution, often called u-substitution. The numerator 3x² + 2x is the derivative of x³ + x² + 4 in the denominator. Set u = x³ + x² + 4, du = (3x² + 2x) dx, simplifying to ∫ du/u. Integration yields ln|u| + C, or ln|x³ + x² + 4| + C. A tempting distractor is choice B, including 1/2 unnecessarily, perhaps from miscalculating the derivative coefficient. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.
In a decay model, compute $\int \frac{7}{2-7x}\,dx$ for the accumulated log-change.
Answer: $-\ln|2-7x|+C$ This integral involves a linear expression in the denominator, requiring careful u-substitution with attention to signs. Let u = 2 - 7x, so du = -7 dx, which means dx = -⅟₇ du. The integral becomes ∫ 7/u · (-⅟₇) du = -∫ du/u = -ln|u| + C = -ln|2-7x| + C. Choice A (ln|2-7x| + C) misses the negative sign that comes from the negative derivative of (2-7x). When the expression inside a logarithm has a negative coefficient on x, the antiderivative includes a negative sign.
A population model needs $\int \dfrac{4}{1+2t}\,dt$ . Find an antiderivative.
Answer: $2\ln|1+2t|+C$ The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is 1 + 2t, whose derivative 2 requires adjusting the 4 to 2. Set u = 1 + 2t, so du = 2 dt and the integral becomes 2 ∫ du/u. Integrating gives 2 ln|u| + C, or 2 ln|1 + 2t| + C. A tempting distractor like 4 ln|1 + 2t| + C fails because it neglects to divide by 2 from du, doubling the coefficient. To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.
A rate integral is $\int \dfrac{6x-8}{\sqrt{3x^2-8x+5}}\,dx$ . Find an antiderivative.
Answer: $2\sqrt{3x^2-8x+5}+C$ This integral requires the skill of integration by substitution, often called u-substitution. The numerator 6x - 8 is the derivative of 3x² - 8x + 5 under the square root. Set u = 3x² - 8x + 5, du = (6x - 8) dx, simplifying to ∫ $u^{-1/2}$ du. Integration yields 2 $u^{1/2}$ + C, or 2 √(3x² - 8x + 5) + C. A tempting distractor is choice A, using 1 instead of 2, possibly from forgetting to multiply by 2 in the power rule. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.
A charge model includes $\int \dfrac{14x}{\sqrt{7x^2+3}}\,dx$ . Find an antiderivative.
Answer: $2\sqrt{7x^2+3}+C$ The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is 7x² + 3, whose derivative 14x exactly matches the numerator. Set u = 7x² + 3, so du = 14x dx, transforming the integral into ∫ du / √u. Integrating gives 2 √u + C, or 2 √(7x² + 3) + C. A tempting distractor like √(7x² + 3) + C fails because it misses the 2 from integrating $u^{-1/2}$ . To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.
Evaluate $\int 4x\sin(x^2)\,dx$ , where the input to sine is $x^2$ .
Answer: $-2\cos(x^2)+C$ This integral uses u-substitution where we need to account for the chain rule carefully. The inner function is $x^2$ with derivative $2x$ , but we have $4x$ in the integrand. Setting $u = x^2$ gives $du = 2x\,dx$ , so $4x\,dx = 2\,du$ . The integral becomes $2\int \sin(u)\,du = -2\cos(u) + C$ . Substituting back yields $-2\cos(x^2) + C$ . Choice E incorrectly divides by 2 instead of multiplying, mishandling the coefficient adjustment. When the coefficient doesn't match exactly, adjust by the ratio of what you have to what you need.
During cooling, compute $\int_0^1 6x\,e^{3x^2}\,dx$ for total heat released over time.
Answer: $e^3-1$ This definite integral requires u-substitution to handle the exponential with a composite exponent. Let u = 3x², so du = 6x dx, which matches the 6x dx in our integrand perfectly. The integral becomes ∫ eᵘ du = eᵘ + C. For the definite integral, when x = 0, u = 0, and when x = 1, u = 3, giving us e³ - e⁰ = e³ - 1. Choice B (⅙(e³-1)) might seem reasonable if you incorrectly think the 6 creates a factor of ⅙, but the 6x is entirely absorbed by the du substitution. Remember that when the derivative of the exponent appears as a factor, u-substitution will cleanly evaluate the integral.
A tank's volume change is modeled by $\int \frac{2x}{1+x^2}\,dx$ ; which antiderivative is correct?
Answer: $\ln(1+x^2)+C$ This integral calls for u-substitution because we have a rational function where the numerator relates to the derivative of the denominator. Setting u = 1 + x², we find du = 2x dx, which means x dx = ½ du. The integral becomes ∫ (2x)/(1+x²) dx = ∫ 2 · (½ du)/u = ∫ du/u = ln|u| + C = ln(1+x²) + C. Choice C (½ln(1+x²) + C) is tempting but incorrectly applies an extra factor of ½, forgetting that the 2 in the numerator cancels with the ½ from the substitution. When you see a fraction where the numerator is nearly the derivative of the denominator, think logarithm and u-substitution.
Compute $\int \dfrac{2x+7}{(x^2+7x+1)^4}\,dx$ .
Answer: $-\dfrac{1}{3(x^2+7x+1)^3}+C$ The skill here is integrating using u-substitution. The denominator is $(x² + 7x + 1)^4$ , and numerator 2x + 7 is its derivative. Set u = x² + 7x + 1 with du = (2x + 7) dx, so ∫ $u^{-4}$ du = -1/( $3u^3$ ) + C. This is -1/[3 $(x² + 7x + 1)^3$ ] + C. Choice D omits the 3 in the denominator, forgetting the power rule division. Identify negative powers in denominators with matching numerators for substitution.
A diffusion model uses $\int \dfrac{(2x-1)}{(x^2-x+6)^3}\,dx$ . Find an antiderivative.
Answer: $-\dfrac{1}{2(x^2-x+6)^2}+C$ This integral requires the skill of integration by substitution, often called u-substitution. The numerator (2x - 1) matches the derivative of the inner quadratic x² - x + 6 in the denominator raised to the third power. Setting u = x² - x + 6 gives du = (2x - 1) dx, simplifying to ∫ $u^{-3}$ du. Integration results in -1/(2u²) + C, or -1/(2(x² - x + 6)²) + C. A tempting distractor is choice C, which omits the factor of 1/2, possibly from forgetting to divide by -2 when integrating $u^{-3}$ . To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.
Compute $\int (3x^2-4x)(x^3-2x^2+7)^5\,dx$ from a composite response function.
Answer: $\dfrac{(x^3-2x^2+7)^6}{6}+C$ The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is x³ - 2x² + 7, whose derivative 3x² - 4x exactly matches the factor outside. Set u = x³ - 2x² + 7, so du = (3x² - 4x) dx, transforming the integral into ∫ u⁵ du. Integrating gives (1/6) u⁶ + C, or (x³ - 2x² + 7)⁶ / 6 + C. A tempting distractor like (x³ - 2x² + 7)⁶ / 3 + C fails because it uses the wrong divisor, possibly confusing the exponent with the coefficient. To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.
Compute $\int \dfrac{5}{(2x-3)^2}\,dx$ for a rate with inverse-square dependence.
Answer: $-\dfrac{5}{2(2x-3)}+C$ The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is 2x - 3, whose derivative 2 requires dividing by 2. Set u = 2x - 3, so du = 2 dx and the integral becomes (5/2) ∫ u⁻² du. Integrating gives (5/2) × (-1/u) + C = -5/(2u) + C, or -5/(2(2x - 3)) + C. A tempting distractor like -5/(2x - 3) + C fails because it omits the division by 2 from du, making the coefficient too large. To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.
Compute $\int \dfrac{2x+4}{(x^2+4x+1)}\,dx$ for $x^2+4x+1\ne 0$ .
Answer: $\ln|x^2+4x+1|+C$ This integral requires the skill of integration by substitution, often called u-substitution. The numerator 2x + 4 is the derivative of x² + 4x + 1 in the denominator. Set u = x² + 4x + 1, du = (2x + 4) dx, transforming to ∫ du/u. This integrates to ln|u| + C, or ln|x² + 4x + 1| + C. A tempting distractor is choice B, multiplying by 1/2, possibly from mistakenly halving the derivative. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.
A mechanics integral is $\int \dfrac{(10x)}{(5x^2+2)^5}\,dx$ . Find an antiderivative.
Answer: $-\dfrac{1}{4(5x^2+2)^4}+C$ This integral requires the skill of integration by substitution, often called u-substitution. The numerator 10x is the derivative of 5x² + 2 in the denominator raised to 5. Set u = 5x² + 2, du = 10x dx, simplifying to ∫ $u^{-5}$ du. Integration gives -1/( $4u^4$ ) + C, or -1/(4 $(5x² + 2)^4$ ) + C. A tempting distractor is choice A, omitting the 1/4, likely from forgetting to divide by -4 in the power rule. To recognize opportunities for u-substitution, look for a composite function where the derivative of the inner part appears as a factor in the integrand.
Compute $\int \dfrac{(2x)}{(x^2+1)}\,dx$ .
Answer: $\ln(x^2+1)+C$ The skill here is integrating using u-substitution. The denominator is x² + 1, with numerator 2x as its derivative. Set u = x² + 1 with du = 2x dx, yielding ∫ du/u = ln|u| + C. This is ln(x² + 1) + C. Choice B halves incorrectly, as the 2 is in du. Look for non-linear denominators with derivative numerators for logarithmic results.
Find $\int \dfrac{(2x+1)}{(x^2+x+4)}\,dx$ representing accumulated logarithmic change.
Answer: $\ln|x^2+x+4|+C$ The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is x² + x + 4, whose derivative 2x + 1 exactly matches the numerator. Set u = x² + x + 4, so du = (2x + 1) dx, transforming the integral into ∫ du/u. Integrating gives ln|u| + C, or ln|x² + x + 4| + C. A tempting distractor like (1/2) ln|x² + x + 4| + C fails because it unnecessarily halves the result, perhaps from misapplying a coefficient to du. To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.
Evaluate $\int \dfrac{(4x-1)}{\sqrt{2x^2-x+9}}\,dx$ .
Answer: $2\sqrt{2x^2-x+9}+C$ The skill here is integrating using u-substitution. The square root contains 2x² - x + 9, with numerator 4x - 1 as its derivative. Set u = 2x² - x + 9 with du = (4x - 1) dx, yielding ∫ du/√u = 2√u + C. This is 2√(2x² - x + 9) + C. Choice C halves instead of doubling, missing the integral of $u^{-1/2}$ . Seek root compositions where numerators equal inner derivatives.
A light sensor output requires $\int \dfrac{6}{(3x-2)}\,dx$ . What is an antiderivative?
Answer: $2\ln|3x-2|+C$ The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is 3x - 2, whose derivative 3 requires adjusting the 6 to 2. Set u = 3x - 2, so du = 3 dx and the integral becomes 2 ∫ du/u. Integrating gives 2 ln|u| + C, or 2 ln|3x - 2| + C. A tempting distractor like 6 ln|3x - 2| + C fails because it doesn't divide by the 3 from du, resulting in too large a coefficient. To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.

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