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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

sin (x^{3})

x \to 0 lim (- 3 x^{3} - 7 x + 8)

\int_{- 2}^{2} 5 d x

y^{^{''}} + 9 y^{^{'}} + 20 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\frac{d y}{d x} = \frac{3 x}{e ^{y}}

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