Evaluate the integral $\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x$

Question

Evaluate the integral $$\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x.$$

If I transfer $(x^2 + 4)^5$ to the numerator, how do I integrate?

Answer 1

I don't know what you mean by "transfer" but:

Hint Note that $d(x^2 + 4) = 2 x \,dx$, so the $x$ in the numerator suggests a particular substitution.

Answer 2

with $u = x^2 + 4$: $$ \int (x^2 + 4)^{-5 } {x dx} = \frac 12 \int u^{-5} {du} = - \frac 1{8} u^{-4} = - \frac 1{8} (x^2 + 4)^{-4} $$

Answer 3

Put $u = x^2+4 \implies du = 2x\,dx\implies x\,dx = \frac 12 du$.

$$ \begin{align}\int \frac{x}{(x^2 + 4)^5} \mathrm{d}x &= \frac 12\int u^{-5} \,du\\ & = \frac 12 \left(\frac {u^{-4}}{-4}\right) +c \\ &= -\frac{1}{8u^4} + c \\ &= -\frac {1}{8(x^2 + 4)^4} + c\end{align} $$