Integrate $\int \frac{16x+16}{(x^2+2x+8)^9}dx$ using u substitution.

Question

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:

$\int \frac{16x+16}{(x^2+2x+8)^9}dx$

$= \int \frac{16x+16}{u^9}\frac{du}{2x+2}$

$= \int \frac{8(2x+2)}{u^9}\frac{du}{2x+2}$

$= \int \frac{8}{u^9}du$

$= 8 \int u^{-9} du$

$= 8 \frac{u^{-8}}{-8}+C$

$= 8 \frac{(x^2+2x+8)^{-8}}{-8}+C$

$= -(x^2+2x+8)^{-8}+C$

Answer 1

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

$$\int\frac{16x+18}{(x^2+2x+8)^9}dx=8\int\overbrace{(2x+2)}^{=(x^2+2x+8)'}(x^2+2x+8)^{-9}dx=$$

$$=8\frac{(x^2+2x+8)^{-8}}{-8}+C=-\frac1{(x^2+2x+8)^8}+C$$

Answer 2

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 $$\frac{\text{d}}{\text{d}x} \left(-(x^2+2x+8)^{-8}+C\right)=-(-8)(x^2+2x+8)^{-9}(2x+2)=$$ $$=8(x^2+2x+8)^{-9}(2x+2)=\frac{16x+16}{(x^2+2x+8)^9}$$

Answer 3

In this case, very conveniently, the numerator is (a multiple of) the derivative of the polynomial being raised to the $-9$- th power.

Thus this function is easy to integrate. Use the power rule and the chain rule.