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

\frac{\partial}{\partial x} (2 x e^{y})

\frac{1}{x}, at x = 3

5 sin^{2} (x^{2})

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

\int_{0}^{5} \frac{5 t}{( t - 6 ) ^{2}} d t

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