Integrating Using Substitution - AP Calculus BC

A model for volume uses $\int \dfrac{2x+1}{(x^2+x+3)}\,dx$ ; evaluate it.
Answer: $\ln\left(x^2+x+3\right)+C$ This integral requires u-substitution for a rational function where the numerator is the derivative of the denominator. Let $u = x^2 + x + 3$ , so $du = (2x + 1) \, dx$ , making the integral $\int \frac{1}{u} \, du = \ln|u| + C = \ln(x^2 + x + 3) + C$ . Since $x^2 + x + 3 = (x + \frac{1}{2})^2 + \frac{11}{4} > 0$ , we can omit absolute value bars. Choice B incorrectly includes a factor of 1/2, while choice D loses the quadratic terms. When the numerator exactly equals the derivative of the denominator, use logarithmic integration.
A periodic motion needs $\int \dfrac{6\sin(3x)}{\cos(3x)}\,dx$ ; evaluate the integral.
Answer: $-2\ln|\cos(3x)|+C$ This integral requires u-substitution for a rational trigonometric function. The integrand 6sin(3x)/cos(3x) = 6tan(3x) can be rewritten, but it's better to use u = cos(3x), so du = -3sin(3x) dx. This gives us ∫6sin(3x)/cos(3x) dx = ∫(-2/u) du = -2ln|u| = -2ln|cos(3x)|. The factor 6 divided by -3 gives -2. Choice B has the wrong sign, while choice E represents the derivative form. When integrating f'(x)/f(x), use u-substitution to get ln|f(x)| + C.
A thermal model uses $\int \dfrac{15}{(5t-1)}\,dt$ ; compute the integral.
Answer: $3\ln|5t-1|+C$ This integral uses u-substitution for a rational function with a linear denominator. Let u = 5t - 1, so du = 5 dt, which means 15 dt = 3 du. The integral becomes ∫(3/u) du = 3 ln|u| + C = 3 ln|5t - 1| + C. The coefficient 15 divided by 5 gives the factor 3. Choice A incorrectly multiplies by 15, while choice C omits the factor 3. When integrating a constant over a linear expression, divide the constant by the coefficient of the linear term.
A system response uses $\int \dfrac{5}{(x^2+1)}\cdot 2x\,dx$ ; compute it.
Answer: $5\ln(x^2+1)+C$ This integral uses u-substitution for a rational function where the numerator is a multiple of the derivative of the denominator. Let u = x² + 1, so du = 2x dx, which means 5(2x) dx = 10x dx becomes ∫(5/u) du = 5 ln|u| + C = 5 ln(x² + 1) + C. Since x² + 1 > 0, we can omit absolute value bars. Choice B uses 5/2 incorrectly, while choice D would result from an arctangent substitution. When the numerator is a multiple of the derivative of the denominator, use logarithmic integration.
A model for oscillation asks for $\int 3\sec^2(3x+4)\,dx$ ; compute it.
Answer: $\tan(3x+4)+C$ This integral uses u-substitution for a trigonometric function with a linear argument. Let u = 3x + 4, so du = 3 dx, making the integral ∫sec²(u) du = tan(u) + C = tan(3x + 4) + C. The antiderivative of sec²(u) is tan(u), and the factor 3 is already present in the integrand. Choice B incorrectly multiplies by 3, while choice C divides by 3 unnecessarily. When integrating 3sec²(3x + 4), the 3's cancel to give tan(3x + 4) + C.
A concentration curve uses $\int \dfrac{2}{(x-1)^2}\,dx$ ; find the antiderivative.
Answer: $-\dfrac{2}{x-1}+C$ This integral uses u-substitution for a function with a linear expression raised to a negative power. Let u = x - 1, so du = dx, making the integral ∫(2/u²) du = 2∫ $u^{-2}$ du = 2(-1/u) = -2/u = -2/(x - 1). The antiderivative of $u^{-2}$ is - $u^{-1}$ = -1/u. Choice B has the wrong sign, while choice E represents the original function rather than its antiderivative. When integrating 1/(x - a)ⁿ where n > 1, the result is negative.
A cost function involves $\int (x^2+4x+5)^3(2x+4)\,dx$ ; find an antiderivative.
Answer: $\dfrac{1}{4}(x^2+4x+5)^4+C$ This integral uses u-substitution for a power function with a quadratic inner expression. Let u = x² + 4x + 5, so du = (2x + 4) dx, making the integral ∫u³ du = u⁴/4 = (1/4)(x² + 4x + 5)⁴. The derivative of the inner function 2x + 4 exactly matches the linear factor in the integrand. Choice B omits the factor of 1/4, while choice C has an incorrect denominator. When the linear factor exactly matches the derivative of a quadratic expression, use u-substitution with the quadratic as u.
In a statistics transform, compute $\int \dfrac{3}{(1+2x)}\,dx$ .
Answer: $\dfrac{3}{2}\ln|1+2x|+C$ This integral uses u-substitution for a rational function with a linear denominator. Let $u = 1 + 2x$ , so $du = 2 \, dx$ , which means $3 \, dx = \frac{3}{2} du$ . The integral becomes $\int \frac{3}{2} \frac{1}{u} \, du = \frac{3}{2} \ln|u| + C = \frac{3}{2} \ln|1 + 2x| + C$ . The coefficient 3 divided by the derivative factor 2 gives 3/2. Choice B incorrectly multiplies by 3 instead of dividing by 2, while choice C omits the adjustment factor. When integrating over a linear expression $ax + b$ , divide the coefficient by a.
A decay model uses $\int 12e^{-6t}\,dt$ ; find an antiderivative.
Answer: $-2e^{-6t}+C$ This integral uses u-substitution for an exponential function with a negative linear exponent. Let u = -6t, so du = -6 dt, which means dt = -(1/6) du. The integral becomes ∫ $12e^u$ (-1/6) du = -2∫ $e^u$ du = - $2e^u$ = - $2e^{-6t}$ . The factor 12 combined with -1/6 gives -2. Choice B has the wrong sign, while choice C incorrectly multiplies by -12. For exponential functions with negative exponents, the antiderivative will be negative times the exponential divided by the coefficient.
A periodic signal uses $\int \dfrac{6\cos(3x)}{\sin(3x)}\,dx$ ; evaluate the integral.
Answer: $2\ln|\sin(3x)|+C$ This integral uses u-substitution for a rational trigonometric function. The integrand can be written as 6cos(3x)/sin(3x) = 6cot(3x), but it's better to use u = sin(3x), so du = 3cos(3x) dx. This gives us ∫(6cos(3x)/sin(3x)) dx = ∫(2/u) du = 2ln|u| = 2ln|sin(3x)|. The factor 6 divided by 3 gives 2. Choice B has the wrong sign, while choice C incorrectly multiplies by 6. When integrating f'(x)/f(x), use u-substitution to get ln|f(x)| + C.
A motion equation uses $\int \dfrac{4}{(2x+1)^3}\,dx$ ; evaluate the integral.
Answer: $-\dfrac{1}{(2x+1)^2}+C$ This integral uses u-substitution for a function with a linear expression raised to a negative power. Let $u = 2x + 1$ , so $du = 2 \, dx$ , which means $4 \, dx = 2 \, du$ . The integral becomes $\int \frac{2}{u^3} \, du = 2 \int u^{-3} \, du = 2 \left( -\frac{1}{2} u^{-2} \right) = - \frac{1}{u^2} = -\frac{1}{(2x + 1)^2}$ . The antiderivative of $u^{-3}$ is $-\frac{1}{2} u^{-2}$ . Choice A doubles the coefficient incorrectly. When integrating powers of linear expressions, substitute and use the power rule with chain rule adjustment.
A signal processing step needs $\int \dfrac{7}{\sqrt{9- x}}\,dx$ ; find the antiderivative.
Answer: $-14\sqrt{9-x}+C$ This integral uses u-substitution for a function with a square root and linear inner expression. Let u = 9 - x, so du = -dx, which means 7 dx = -7 du. The integral becomes ∫(7/√u)(-1) du = -7∫ $u^{-1/2}$ du = -7(2√u) = -14√u = -14√(9 - x). The antiderivative of $u^{-1/2}$ is $2u^{1/2}$ , and we include the negative from the substitution. Choice B omits the negative sign that comes from the chain rule. When the inner function has a negative derivative, include the negative sign in the result.
A tank's inflow rate is modeled by $r(t)=\dfrac{6t}{t^2+4}$ ; find $\int \dfrac{6t}{t^2+4}\,dt$ .
Answer: $3\ln\left(t^2+4\right)+C$ This integral requires u-substitution to handle the rational function with a quadratic denominator. Let u = t² + 4, so du = 2t dt, which means 6t dt = 3 du. The integral becomes ∫(3/u) du = 3 ln|u| + C. Substituting back gives 3 ln(t² + 4) + C (absolute value bars can be dropped since t² + 4 > 0). Choice D might tempt students who think of arctangent formulas, but that applies to ∫1/(t² + 4) dt, not our integral with t in the numerator. When the numerator contains the derivative of the denominator (up to a constant), use u-substitution with the denominator as u.
A dynamics problem asks for $\int \dfrac{(4x+3)}{\sqrt{2x^2+3x+1}}\,dx$ .
Answer: $2\sqrt{2x^2+3x+1}+C$ This integral requires u-substitution for a function involving a square root with a quadratic inner expression. Let u = 2x² + 3x + 1, so du = (4x + 3) dx, making the integral ∫(1/√u) du = 2√u = 2√(2x² + 3x + 1). The antiderivative of $u^{-1/2}$ is $2u^{1/2}$ . Choice B omits the factor of 2, while choice C doubles it incorrectly. When the numerator is exactly the derivative of the expression under the square root, integrate to get twice the square root.
A model for resistance needs $\int \dfrac{7}{\sqrt{5x+1}}\,dx$ ; compute the antiderivative.
Answer: $\dfrac{14}{5}\sqrt{5x+1}+C$ This integral uses u-substitution for a function with a square root and linear inner expression. Let u = 5x + 1, so du = 5 dx, which means 7 dx = (7/5) du. The integral becomes ∫(7/5)(1/√u) du = (7/5)∫ $u^{-1/2}$ du = (7/5)(2√u) = (14/5)√u = (14/5)√(5x + 1). The antiderivative of $u^{-1/2}$ is $2u^{1/2}$ . Choice B omits the factor of 2 from the power rule. When integrating 1/√(ax + b), substitute u = ax + b and divide by a.
In a cooling law, find $\int 7e^{3t+2}\,dt$ .
Answer: $\dfrac{7}{3}e^{3t+2}+C$ This integral uses u-substitution for an exponential function with a linear exponent. Let u = 3t + 2, so du = 3 dt, which means dt = (1/3) du. The integral becomes ∫7eᵘ(1/3) du = (7/3)∫eᵘ du = (7/3)eᵘ = (7/3) $e^{3t+2}$ . The chain rule requires dividing by the derivative of the exponent, which is 3. Choice A forgets this adjustment factor, while choice E loses the constant 2 in the exponent. For exponential functions with linear exponents ax + b, substitute u = ax + b and divide by the coefficient a.
A kinematics substitution arises in $\int (1+\sin x)\cos x\,dx$ ; find an antiderivative.
Answer: $\dfrac{1}{2}(1+\sin x)^2+C$ This integral uses u-substitution for a trigonometric function where we can identify the inner function. Let u = 1 + sin x, so du = cos x dx, making the integral ∫u du = u²/2 = (1/2)(1 + sin x)². This expands to (1/2)(1 + 2sin x + sin²x), which includes both linear and quadratic sine terms. Choice B omits the factor of 1/2 from integration, while choice C incorrectly substitutes cosine. When integrating a function times the derivative of an inner function, substitute the inner function.
A control system uses $\int \dfrac{6}{(3x+2)}\,dx$ ; compute the integral.
Answer: $2\ln|3x+2|+C$ This integral uses u-substitution for a rational function with a linear denominator. Let u = 3x + 2, so du = 3 dx, which means 6 dx = 2 du. The integral becomes ∫(2/u) du = 2 ln|u| + C = 2 ln|3x + 2| + C. The coefficient 6 divided by the derivative factor 3 gives 2. Choice A incorrectly multiplies by 6, while choice C omits the adjustment factor. When integrating a constant over a linear expression ax + b, divide the constant by a.
In modeling charge, compute $\int \dfrac{6}{(x-4)}\,dx$ .
Answer: $6\ln|x-4|+C$ This integral uses u-substitution for a simple rational function with a linear denominator. Let u = x - 4, so du = dx, making the integral ∫(6/u) du = 6 ln|u| + C = 6 ln|x - 4| + C. The constant 6 remains as a coefficient in front of the natural logarithm. Choice A omits the coefficient 6, while choice C incorrectly divides by 6. When integrating a constant over a linear expression of the form x - a, the result is the constant times ln|x - a| + C.
An optics computation requires $\int \dfrac{9x^2}{x^3+7}\,dx$ ; find an antiderivative.
Answer: $3\ln|x^3+7|+C$ This integral requires u-substitution for a rational function where the numerator is a multiple of the derivative of the denominator. Let u = x³ + 7, so du = 3x² dx, which means 9x² dx = 3 du. The integral becomes ∫(3/u) du = 3 ln|u| + C = 3 ln|x³ + 7| + C. The factor 9 divided by 3 gives the coefficient 3. Choice A incorrectly uses 9, while choice C omits the factor entirely. When the numerator is a constant multiple of the derivative of the denominator, use logarithmic integration.

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