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

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

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

integral

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

\frac{c}{r ^{3}}

\frac{sec ( x )}{tan ^{2} ( x )}

(x - 5) ln (x)

x^{5} - x + 1

- 4 ln (x)

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