What is the integral of x/((x^2+2x+2)^2) ?

The integral of x/((x^2+2x+2)^2) is -1/(2(x^2+2x+2))-1/2 (arctan(x+1)+(x+1)/(x^2+2x+2))+C

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integral of x/((x^2+2x+2)^2)

\int \frac{x}{( x ^{2} + 2 x + 2 ) ^{2}} d x

- \frac{1}{2 ( x ^{2} + 2 x + 2 )} - \frac{1}{2} (arctan (x + 1) + \frac{x + 1}{x ^{2} + 2 x + 2}) + C

Solution steps

\int \frac{x}{( x ^{2} + 2 x + 2 ) ^{2}} d x

Complete the square

x^{2} + 2 x + 2 : (x + 1)^{2} + 1

= \int \frac{x}{( ( x + 1 ) ^{2} + 1 ) ^{2}} d x

Apply u-substitution

= \int \frac{u - 1}{( u ^{2} + 1 ) ^{2}} d u

Expand

\frac{u - 1}{( u ^{2} + 1 ) ^{2}} : \frac{u}{( u ^{2} + 1 ) ^{2}} - \frac{1}{( u ^{2} + 1 ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{u}{( u ^{2} + 1 ) ^{2}} d u - \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u

\int \frac{u}{( u ^{2} + 1 ) ^{2}} d u = - \frac{1}{2 ( u ^{2} + 1 )}

\int \frac{1}{( u ^{2} + 1 ) ^{2}} d u = \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}})

= - \frac{1}{2 ( u ^{2} + 1 )} - \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}})

Substitute back

u = x + 1

= - \frac{1}{2 ( ( x + 1 ) ^{2} + 1 )} - \frac{1}{2} (arctan (x + 1) + \frac{x + 1}{1 + ( x + 1 ) ^{2}})

Simplify

- \frac{1}{2 ( ( x + 1 ) ^{2} + 1 )} - \frac{1}{2} (arctan (x + 1) + \frac{x + 1}{1 + ( x + 1 ) ^{2}}) : - \frac{1}{2 ( x ^{2} + 2 x + 2 )} - \frac{1}{2} (arctan (x + 1) + \frac{x + 1}{x ^{2} + 2 x + 2})

= - \frac{1}{2 ( x ^{2} + 2 x + 2 )} - \frac{1}{2} (arctan (x + 1) + \frac{x + 1}{x ^{2} + 2 x + 2})

Add a constant to the solution

= - \frac{1}{2 ( x ^{2} + 2 x + 2 )} - \frac{1}{2} (arctan (x + 1) + \frac{x + 1}{x ^{2} + 2 x + 2}) + C

What is the integral of x/((x^2+2x+2)^2) ?
The integral of x/((x^2+2x+2)^2) is -1/(2(x^2+2x+2))-1/2 (arctan(x+1)+(x+1)/(x^2+2x+2))+C