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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

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