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

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

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integral of (2x^2)/((x^2+1)^2)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

Apply the Sum Rule:

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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