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

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

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

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

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

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Take the partial fraction of

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

= \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

= \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}})

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{x ^{a}} d x

x \to - 1 + lim (\frac{x}{x + 1})

\frac{\partial}{\partial t} (3 t)

\frac{d y}{d x} = \frac{x + y - 2}{2 x + 2 y + 3}

\int \frac{1}{x ^{2} - 25} d x

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