Integrate $\int \frac{16x+16}{(x^2+2x+8)^9}dx$ using u substitution. Section 5.2
Can somebody verify this solution for me?
Integrate $\int \frac{16x+16}{(x^2+2x+8)^9}dx$ using u substitution.
Let $u=x^2+2x+8$. Then $\frac{du}{dx}=2x+2$ and so $\frac{du}{2x+2}=dx$. Thus we have:
$\int \frac{16x+16}{(x^2+2x+8)^9}dx$
$= \int \frac{16x+16}{u^9}\frac{du}{2x+2}$
$= \int \frac{8(2x+2)}{u^9}\frac{du}{2x+2}$
$= \int \frac{8}{u^9}du$
$= 8 \int u^{-9} du$
$= 8 \frac{u^{-8}}{-8}+C$
$= 8 \frac{(x^2+2x+8)^{-8}}{-8}+C$
$= -(x^2+2x+8)^{-8}+C$

You don't need online calculators to check your answers in indefinite integrals, the best thing you can learn is to derive again your result and hoping it will give you back the integrand function (of course when the calculations aren't too much brutal).
Indeed
$$\frac{\text{d}}{\text{d}x} \left(-(x^2+2x+8)^{-8}+C\right)=-(-8)(x^2+2x+8)^{-9}(2x+2)=$$
$$=8(x^2+2x+8)^{-9}(2x+2)=\frac{16x+16}{(x^2+2x+8)^9}$$
