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

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\frac{\partial}{\partial y} (4 x y^{3} (z^{2} + 1)^{3})

f (x) = \frac{e ^{3 x}}{x ^{15}}

x \to + 2 - lim (x - 2)

\int \frac{e ^{u}}{( 8 - e ^{u} ) ^{2}} d u

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