Question

∫(x^{2}+4)^{3} }" alt="\frac{xdx}{(x^{2}+4)^{3} }" align="absmiddle" class="latex-formula">

Answer 1

Substitute $u=x^2+4$ and $du=2x\,dx$. Then the integral transforms to

$\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \int \frac{du}{u^3}$

Apply the power rule.

$\displaystyle \int \frac{du}{u^3} = -\dfrac1{2u^2} + C$

Now put the result back in terms of $x$.

$\displaystyle \int \frac{x\,dx}{(x^2+4)^3} = \frac12 \left(-\dfrac1{2u^2} + C\right) = -\dfrac1{4u^2} + C = \boxed{-\dfrac1{4(x^2+4)^2} + C}$