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

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

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

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

2 arctan (y) - y + C

Solution steps

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

\frac{1 - y ^{2}}{1 + y ^{2}} = \frac{2}{1 + y ^{2}} - 1

= \int \frac{2}{1 + y ^{2}} - 1 d y

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 + y ^{2}} d y - \int 1 d y

\int \frac{2}{1 + y ^{2}} d y = 2 arctan (y)

\int 1 d y = y

= 2 arctan (y) - y

Add a constant to the solution

= 2 arctan (y) - y + C

\int 1.2 x d x

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

\int \frac{3 x ^{5}}{2} d x

\int (x + y)^{3} d y

\int \frac{in ( x + 1 )}{x + 1} d x

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