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

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

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

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

- 2 arctan (x) + x + C

Solution steps

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

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

\int 2 x (x^{2} + 1)^{3} d x

\int sec^{3} (2 x) tan (2 x) d x

x^{2} (1 - x)^{2}

\int x^{2} e^{- 2 x^{2}} d x

y = 1 - x^{2}, y = 3 - 3 x

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