How do you integrate $\int \frac{1}{{(x^{2} + 2 x + 2)}^{2}}$ $int 1/(x^2+2x+2)^2$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{{(x^{2} + 2 x + 2)}^{2}}$ $int 1/(x^2+2x+2)^2$ by trigonometric substitution?

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Answer 1 · 2016-12-23T21:34:51

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

Explanation:

Substitute $w = {tan}^{- 1} (x + 1)$ $w=tan^-1(x+1)$ , $\frac{d w}{d x} = \frac{1}{x^{2} + 2 x + 2}$ $(dw)/dx=1/(x^2+2x+2)$ .
The integral becomes $\int {cos}^{2} w d w$ $int cos^2w dw$
$= \int \frac{1 + cos 2 w}{2} d w$ $=int(1+cos 2w)/2 dw$
$= \frac{w}{2} + (\frac{1}{2}) sin w cos w$ $=w/2+(1/2)sin w cos w$
$= (\frac{1}{2}) {tan}^{- 1} (x + 1) + (\frac{1}{2}) . \frac{x + 1}{\sqrt{x^{2} + 2 x + 2}} \times \frac{1}{\sqrt{x^{2} + 2 x + 2}} + C$ $=(1/2)tan^-1(x+1)+(1/2).(x+1)/sqrt(x^2+2x+2)xx1/sqrt(x^2+2x+2)+C$
$= (\frac{1}{2}) {tan}^{- 1} (x + 1) + (\frac{1}{2}) \frac{x + 1}{x^{2} + 2 x + 2} + C$ $=(1/2)tan^-1(x+1)+(1/2)(x+1)/(x^2+2x+2)+C$

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

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

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How do you integrate $\int \frac{1}{{(x^{2} + 2 x + 2)}^{2}}$ $int 1/(x^2+2x+2)^2$ by trigonometric substitution?

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Explanation:

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How do you integrate ∫1(x2+2x+2)2int 1/(x^2+2x+2)^2 by trigonometric substitution?

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How do you integrate $\int \frac{1}{{(x^{2} + 2 x + 2)}^{2}}$ $int 1/(x^2+2x+2)^2$ by trigonometric substitution?