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

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

Popular >

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{2 x}{x ^{2} + 1} d x = ln x^{2} + 1

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

Add a constant to the solution

= x arctan (\frac{2 x}{1 - x ^{2}}) - ln x^{2} + 1 + C

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

\frac{\partial}{\partial y} (\frac{y}{7 x})

\frac{d}{d x} (\frac{4 e ^{x}}{cos ( x )})

\int_{1}^{4} (x) d x

\frac{d y}{d x} - \frac{y}{x} = 6 x e^{x}

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