# How do you integrate int 1/(x^2+2x+2)^2 by trigonometric substitution?

Dec 23, 2016

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

#### Explanation:

Substitute $w = {\tan}^{-} 1 \left(x + 1\right)$, $\frac{\mathrm{dw}}{\mathrm{dx}} = \frac{1}{{x}^{2} + 2 x + 2}$.
The integral becomes $\int {\cos}^{2} w \mathrm{dw}$
$= \int \frac{1 + \cos 2 w}{2} \mathrm{dw}$
$= \frac{w}{2} + \left(\frac{1}{2}\right) \sin w \cos w$
$= \left(\frac{1}{2}\right) {\tan}^{-} 1 \left(x + 1\right) + \left(\frac{1}{2}\right) . \frac{x + 1}{\sqrt{{x}^{2} + 2 x + 2}} \times \frac{1}{\sqrt{{x}^{2} + 2 x + 2}} + C$
$= \left(\frac{1}{2}\right) {\tan}^{-} 1 \left(x + 1\right) + \left(\frac{1}{2}\right) \frac{x + 1}{{x}^{2} + 2 x + 2} + C$

It is helpful to draw a right triangle with angle ${\tan}^{-} 1 \left(x + 1\right)$, adjacent side $1$, opposite side $x + 1$ and hypotenuse ${x}^{2} + 2 x + 2$. And remember that $\cos 2 w = 2 {\cos}^{2} w - 1$, $\sin 2 w = 2 \sin w \cos w$.

If the initial substitution is hard to spot, substitute $u = x + 1$ first in order to get rid of the $2 x$, giving $\int \frac{\mathrm{du}}{{u}^{2} + 1} ^ 2$ then further substitute $u = \tan w$, and draw a triangle with sides $1$, $u$, $\sqrt{1 + {u}^{2}}$