# Integral which must be solved using integration by parts

I have to solve this problem using integration by parts. I am new to integration by parts and was hoping someone can help me.

$$\int\frac{x^3}{(x^2+2)^2} dx$$

Here is what I have so far:

$$\int udv = uv-\int vdu$$

$$u=x^2+2$$ Therefore, $$xdx=\frac{du}{2}$$ $$dv=x^3$$ Therefor, $$v=3x^2$$

• so substitution isn't allowed? Jul 13, 2015 at 0:36
• @randomgirl we have to use integration by parts Jul 13, 2015 at 0:40

In integration by parts, you want $v$ to be something that is easy to integrate, $u$ is easy to differentiate, and $v\,du$ is easier to integrate than the original $u\,dv$.

In this case, the difficulty is integrating the fraction; and to make things more difficult, the fraction's denominator is squared. Here's one way to make that easier.

Let $dv=\frac{2x}{(x^2+2)^2}$. That is easy to integrate (can you see it?), and integrating it will reduce the power of the denominator. So then $u=\frac{x^2}2$ to get $u\,dv$ what we want.

Here is the rest:

We can see by inspection that the numerator of $dv$ is the derivative of the denominator. We therefore get $v=-\frac 1{x^2+2}$. Clearly $du=x\,dx$. So,

\begin{align} \int\frac{x^3}{(x^2+2)^2}\,dx &= uv-\int v\,du \\[2ex] &= \frac{x^2}2\cdot -\frac 1{x^2+2}-\int-\frac 1{x^2+2}\cdot x\,dx \\[2ex] &= -\frac 12\frac{x^2}{x^2+2}+\frac 12\int\frac{2x}{x^2+2}\,dx \\[2ex] &= -\frac 12\frac{x^2}{x^2+2}+\frac 12\ln|x^2+2|+C_1 \\[2ex] &= \frac{1}{x^2+2}-\frac 12+\frac 12\ln|x^2+2|+C_1 \\[2ex] &= \frac{1}{x^2+2}+\frac 12\ln|x^2+2|+C_2 \end{align}

Either the last line or the one two lines above it are good answers. Those two lines have different arbitrary constants, since the $-\frac 12$ disappeared between them.

Is that clear?

• can you take a look and see if my answer is correct below? you have helped me in the past. $$\frac{-1}{2}\frac{x^2}{x^2+1}-(\frac{-1}{4}ln(|x^2+2|)$$ Jul 13, 2015 at 14:53
• yes please. I would like to know where I went wrong Jul 13, 2015 at 15:24
• @Csci319: Almost, one answer is $\frac{1}{x^2+2}+\frac 12\ln|x^2+2|+C$, another is $-\frac 12\frac{x^2}{x^2+2}+\frac 12\ln|x^2+2|+C$. Do you need me to show more steps in my answer? Jul 13, 2015 at 15:28
• please, it would help a lot Jul 13, 2015 at 15:34

A couple of things with your work so far: When doing integration by parts, $u$ and $v$ have to be two functions which are multiplied by one another inside the integral. With your choice of $u$ and $v$, this is not the case. Second, to go from $dv$ to $v$, you should integrate, not differentiate. As a hint for how to proceed, you might try $$u=x^2$$ and $$dv=\frac{x}{(x^2+2)^2}dx.$$ Then $$du=2xdx$$ and $$v=-\frac 12\frac{1}{x^2+2}.$$ Can you take it from there?

• so the integral which I need to solve will look like this: $$\int\frac{\frac{-1}{2}}{x^2+2}x^2$$? Jul 13, 2015 at 0:52
• Not quite. The integral you need to solve is $$\int v\,du=\frac{-1}{2}\int\frac{x}{x^2+2}dx.$$ Jul 13, 2015 at 1:09
• so then that would equal $$\frac{-1}{4}ln(|x^2+2|)+C$$? Jul 13, 2015 at 2:09
• Yes, but don't forget to add on $$uv=-\frac{1}{2}\frac{x^2}{x^2+1}.$$ Jul 13, 2015 at 2:19
• so $$\frac{-1}{2}\frac{x^2}{x^2+1}-(\frac{-1}{4}ln(|x^2+2|))+C$$? for the final answer Jul 13, 2015 at 2:46

Let $dv=x/(x^2+2)^2$. and $u=x^2$. $v=\frac{-1/2}{x^2+2}$ and $du=2xdx$. You can now do the integral $\int vdu$ on your own, as it's just a logorithm.

• so the integral which I need to solve will look like this: $$\int\frac{\frac{-1}{2}}{x^2+2}x^2$$? Jul 13, 2015 at 0:52
• that $x^2$ should be a $2x$ right? $\int udv = uv - \int vdu$. Jul 13, 2015 at 1:05

Hint:

$\int\frac{x^3}{(x^2+2)^2}dx$

Write $x^3asx^2x$

$\int\frac{x^2x}{(x^2+2)^2}dx$

• add and substract 2 in numerator

$\int\frac{[(x^2+2)-2]x}{(x^2+2)^2}dx$

• separate it as two integrals

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

$=>\int\frac{x}{(x^2+2)}dx-\int\frac{2x}{(x^2+2)^2}dx$

formulae :(1) $\int\frac{f{'}(x)}{f(x)}dx=log|f(x)|$ formulae:(2)$int\frac{f{'}(x)}{f^{2}(x)}dx=-\frac{1}{f(x)}$

-take $f(x) = x^2+2$ for the first part of integral and before that multiply and divide with 2 for using formulae(1)

• take $f(x) = x^2+2$for the second integral using formulae ( 2)

$=\frac{1}{2}log|x^2+2|+\frac{1}{x^2+2}$
• Your answer is not what I would call a "hint". It is a walkthrough to solving the answer. Also, when you're writing equations, consider using \$\$math\$\$ instead of a single dollar sign. Jul 13, 2015 at 5:36