# How do you find the integral int (x+1)/(x^2+2x+2)^3dx using substitution?

Dec 15, 2016

The answer is $= - \frac{1}{4 \left({x}^{2} + 2 x + 2\right)} + C$

#### Explanation:

We use $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne 0\right)$

Let $u = {x}^{2} + 2 x + 2$

$\mathrm{du} = \left(2 x + 2\right) \mathrm{dx}$

$\left(x + 1\right) \mathrm{dx} = \frac{\mathrm{du}}{2}$

$\int \frac{\left(x + 1\right) \mathrm{dx}}{{x}^{2} + 2 x + 2} ^ 3 = \frac{1}{2} \int \frac{\mathrm{du}}{{u}^{3}} = \frac{1}{2} \int {u}^{- 3} \mathrm{du}$

$= \frac{1}{2} \cdot {u}^{- 3 + 1} / \left(- 3 + 1\right)$

$= \frac{1}{2} \cdot {u}^{- 2} / \left(- 2\right)$

$= - \frac{1}{4 {u}^{2}}$

$= - \frac{1}{4 \left({x}^{2} + 2 x + 2\right)} + C$

Dec 15, 2016

$- \frac{1}{4 {\left({x}^{2} + 2 x + 2\right)}^{2}} + C$

#### Explanation:

You first decide which part of one looks like the other. If you make
$U = {x}^{2} + 2 x + 2$
$\mathrm{dU} = 2 x + 2$
Therefore, all you need to do is multiply the top by 2 and you have the same thing as your dU.
You then have multiply the inside by 2 but also the outside by 1/2.
Once you have this, you can then substitute your U and dU in your situation.
$\frac{1}{2} {\int}_{.} \frac{1}{{\left(U\right)}^{3}} \mathrm{du}$

you final recieve after your done
$\frac{1}{2} \left(- \frac{1}{2 {u}^{2}}\right)$

Then you finally substitute your U back in the situation and you have

$- \frac{1}{4 {\left({x}^{2} + 2 x + 2\right)}^{2}} + C$