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

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

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

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

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

Solution steps

\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

\frac{d}{d x} (- \frac{sin ( ln ( x ) )}{x})

\frac{d}{d x} (- x^{2} y)

d er i v a t i v e 3 - 6 x

y^{^{''}} - 2 y^{^{'}} + y = 6 e^{x}

h \to 0 lim (\frac{3 ^{h} - 1}{h})

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