# Question #60ae7

##### 1 Answer
Sep 27, 2017

After applying some imperative properties from integral, you can find: $f \left(x\right) = - \frac{1}{6} \cdot {\left(\frac{1}{x} + 2\right)}^{6}$

#### Explanation:

You shall use the substitution method. Essentially it says that we can solve some complex integrals just by replacing some functions inside the integral by provisory variables.

We have our problem, starting point:

$\int {x}^{- 2} {\left(\frac{1}{x} + 2\right)}^{5} \mathrm{dx}$

Set:
$u \left(x\right) = \frac{1}{x} + 2$

So we have:

$\int {x}^{- 2} u {\left(x\right)}^{5} \mathrm{dx}$

However, the integral is still not easy to use, even harder and cumbersome. We can go on:

Recalling:

$u \left(x\right) = \frac{1}{x} + 2$

let's find the derivative in function of x:

$\frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{x} ^ 2$ (if you cannot see it, let me know! it is basic derivative properties)

Let's go on more:

$\frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{x} ^ 2$

$\mathrm{dx} = - {x}^{2} \mathrm{du}$

Let's come back to the integral (the one with u instead of x):

$\int {x}^{- 2} u {\left(x\right)}^{5} \mathrm{dx}$
$\int {x}^{- 2} u {\left(x\right)}^{5} \left(- {x}^{2} \mathrm{du}\right)$

By elementary arragements...

$- \int u {\left(x\right)}^{5} \mathrm{du}$

Apply the property: $\int {x}^{n} \mathrm{du} = {x}^{n + 1} / \left(n + 1\right)$

$\int u {\left(x\right)}^{5} \mathrm{du} = - {u}^{6} / 6 = s \left(x\right)$

Bringing back the x from: $u \left(x\right) = \frac{1}{x} + 2$

$s \left(x\right) = - \frac{1}{6} {\left(\frac{1}{x} + 2\right)}^{6}$

PS. It can be shown, but we shall skip it here, that if you find the derivative of s(x), you come back to the starting point, as it should be. The two solutions herein are the same, just apply the binomial expansion to the solution of Eric Sia (to whom I thank for the fast solution, in matter of minutes). I had to find an easier way because Yara was unable to follow the previous solution, because of the binomial expansion.