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

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

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

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

2 arctan (x) - x + C

Solution steps

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

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

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

\int \frac{2}{x ^{2} + 1} 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 (1-x^2)/(x^2+1) ?
The integral of (1-x^2)/(x^2+1) is 2arctan(x)-x+C