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))

integral

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

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

\frac{3}{( x ^{2} + 1 )}

1 3^{x} + 1

arctan (x + 1)

\frac{1}{s ^{3}}

x^{2} + 8 x - 3

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