$\int \frac{x+7}{(x+8)^5}\ dx$

Question

Find $$\int \frac{x+7}{(x+8)^5}\ dx$$

I tried $\int \frac{x+7}{(x+8)^5}\ dx$ but I couldn’t complete it. Is there any simpler way to integrate this?

Answer 1

$$\frac{x+7}{(x+8)^5}=\frac{x+8-1}{(x+8)^5}=\frac{x+8}{(x+8)^5}-\frac{1}{(x+8)^5}=\frac{1}{(x+8)^4}-\frac{1}{(x+8)^5}.$$

Thus, we get $$\frac{(x+8)^{-3}}{-3}-\frac{(x+8)^{-4}}{-4}+C.$$ Can you end it now?

Answer 2

Try u-substitution.

Let $u= x+8$ (or $u-1 = x+7$), then $du = (x+8)' dx = dx$ and so we have

$$\int \frac{x+7}{(x+8)^5} \,dx = \int \frac{u-1}{u^5} \, du = \int \frac{1}{u^4} \, du - \int \frac{1}{u^5} \, du$$

Use power rule to solve these integrals and don't forget to substitute $x+8$ back into $u$ when done.

Answer 3

$$\int \frac {x+7}{(x+8)^5} dx$$

Upon the substitution $ u=x+8$ , $du=dx$, we get $$\int \frac {u-1}{u^5} du=$$ $$\int {u^{-4}}-{u^{-5}} du=$$

$$\frac {u^{-3}}{-3} + \frac {u^{-4}}{4} +C=$$

$$\frac {(x+8)^{-3}}{-3} + \frac {(x+8)^{-4}}{4} +C.$$