<s>[INST] 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:
$\int \frac{16x+16}{(x^2+2x+8)^9}dx$
$= \int \frac{16x+16}{u^9}\frac{du}{2x+2}$
$= \int \frac{8(2x+2)}{u^9}\frac{du}{2x+2}$
$= \int \frac{8}{u^9}du$
$= 8 \int u^{-9} du$
$= 8 \frac{u^{-8}}{-8}+C$
$= 8 \frac{(x^2+2x+8)^{-8}}{-8}+C$
$= -(x^2+2x+8)^{-8}+C$
 [/INST] Your solution is correct. You can also use the fact that
$$\int f'(x)\,f(x)^{-9}\,dx=\frac{f(x)^{-8}}{-8}+C$$
and thus
$$\int\frac{16x+18}{(x^2+2x+8)^9}dx=8\int\overbrace{(2x+2)}^{=(x^2+2x+8)'}(x^2+2x+8)^{-9}dx=$$
$$=8\frac{(x^2+2x+8)^{-8}}{-8}+C=-\frac1{(x^2+2x+8)^8}+C$$
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