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

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

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

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

- 2 arctan (x) + x + C

Solution steps

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

\frac{x ^{2} - 1}{1 + x ^{2}} = - \frac{2}{1 + x ^{2}} + 1

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

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

\int 1 d x = x

= - 2 arctan (x) + x

Add a constant to the solution

= - 2 arctan (x) + x + C

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