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, $$v=3x^2$$
 A: 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?
A: 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?
A: 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. 
A: 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)


Then answer is 
$=\frac{1}{2}log|x^2+2|+\frac{1}{x^2+2}$
