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

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

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

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

2 arctan (x) - x + C

Solution steps

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

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

\int 4 x - 15 x^{2} + e^{x} d x

\frac{\partial}{\partial y} (2 x^{3})

\int \frac{4 x}{( x - 1 ) ^{2} ( x + 1 )} d x

\frac{d}{d x} ((x^{2} + 3)^{5})

f (x) = (sin (x))^{2}

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