Integration by Parts

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GRE Quantitative Reasoning › Integration by Parts

Questions 1 - 4

1

Evaluate the following integral.

$\int x e^{4x}dx$

$\frac{x}{4}e^{4x}-\frac{1}{16}e^{4x}+c$

$\frac{x}{4}e^{4x} +\frac{1}{16}e^{4x}+c$

$\frac{x}{4}e^{4x}-\frac{1}{2}x^2+c$

$xe^{4x}+\frac{1}{4}e^{4x}+c$

Explanation

Integration by parts follows the formula:

$\int u \ dv=uv-\int v\ du$

In this problem we have $\int x e^{4x}dx$ so we'll assign our substitutions:

and $dv=e^{4x} dx$

which means and $v=\frac{1}{4}e^{4x}$

Including our substitutions into the formula gives us:

$x\frac{1}{4}e^{4x}-\int \frac{1}{4}e^{4x}dx$

We can pull out the fraction from the integral in the second part:

$\frac{x}{4}e^{4x}-\frac{1}{4} \int e^{4x}dx$

Completing the integration gives us:

$\frac{x}{4}e^{4x}-\frac{1}{4} \left(\frac{1}{4}e^{4x}\right) +c$

$\frac{x}{4}e^{4x}-\frac{1}{16} e^{4x} +c$

2

Integrate the following.

$\int x \cos x \ dx$

$x \sin x + \cos x +c$

$x\sin x-\cos x + c$

$x\sin x+ \sin x + c$

$x\cos x + \sin x + c$

Explanation

Integration by parts follows the formula:

$\int u \ dv=uv- \int v \ du$

So, our substitutions will be and $dv=\cos x \ dx$

which means and $v=\sin x$

Plugging our substitutions into the formula gives us:

$x\sin x-\int \sin x \ dx$

Since $\int \sin x \ dx=-\cos x + c$ , we have:

$x\sin x - (- \cos x + c)$ , or

$x\sin x + \cos x + c$

3

Evaluate the following integral.

$\int x^2 \ln x \ dx$

$\frac{x^{3}}{3}\ln x-\frac{1}{9}x^3+c$

$\frac{x^3}{3}\ln x + c$

$\frac{x^{3}}{3}\ln x+\frac{1}{9}x^3+c$

$\frac{x^{3}}{3}\ln x-\frac{1}{3}x^3+c$

Explanation

Integration by parts follows the formula:

$\int u \ dv=uv-\int v \ du$

Our substitutions will be $u=\ln x$ and

which means $du= \frac{1}{x}dx$ and $v=\frac{1}{3}x^3$ .

Plugging our substitutions into the formula gives us:

$\frac{x^3}{3} \ln x -\int \frac{x^3}{3} \cdot \frac{1}{x} dx$

Look at the integral: we can pull out the $\frac{1}{3}$ and simplify the remaining $\frac{x^3}{x}$ as

$\frac{x^3}{3} \ln x -\frac{1}{3} \int x^2 dx$ .

We now solve the integral: $\int x^2 dx = \frac{1}{3} x^3 + c$ , so:

$\frac{x^3}{3} \ln x -\frac{1}{3} \left(\frac{1}{3}x^3 + c\right)$

$\frac{x^3}{3} \ln x - \frac{1}{9} x^3+c$

4

Evaluate the following integral.

$\int x \cos (2x+1)dx$

$\frac{1}{2}x \sin (2x+1)+ \frac{1}{4}\cos (2x+1)+c$

$\frac{1}{2}x \sin (2x+1)$

$\frac{1}{2}x \sin (2x+1)+ \frac{1}{2}\cos (2x+1)+c$

$\frac{1}{2}x \sin (2x+1)- \frac{1}{2}\cos (2x+1)+c$

Explanation

Integration by parts follows the formula:

$\int u \ dv=uv - \int v \ du$ .

Our substitutions are and $dv= \cos (2x+1) dx$

which means and $v=\frac{1}{2} \sin (2x+1)$ .

Plugging in our substitutions into the formula gives us

$\frac{x}{2} \sin (2x+1)-\int \frac{1}{2} \sin (2x+1) dx$

We can pull $\frac{1}{2}$ outside of the integral.

$\frac{x}{2} \sin (2x+1)-\frac{1}{2} \int \sin (2x+1) dx$

Since $\int \sin (2x+1) dx =-\frac{1}{2} \cos (2x+1)+ c$ , we have

$\frac{x}{2} \sin (2x+1)- \frac{1}{2} \left(-\frac{1}{2} \cos (2x+1) + c\right)$

$\frac{x}{2} \sin (2x+1)+\frac{1}{4} \cos (2x+1) + c$

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