<s>[INST] 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, $$v=3x^2$$
 [/INST] 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 
$=\frac{1}{2}log|x^2+2|+\frac{1}{x^2+2}$
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