Solve ∫ tan^-1y/1+y^2

$$\int \frac { \tan ^ { - 1 } y } { 1 + y ^ { 2 } }$$

Evaluate

$$\frac{x\arctan(y)}{y^{2}+1}+С$$

Solution Steps

Find the integral of \frac{\arctan(y)}{1+y^{2}} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{\arctan(y)}{1+y^{2}}x$$

Simplify.

$$\frac{\arctan(y)x}{1+y^{2}}$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{\arctan(y)x}{1+y^{2}}+С$$

Differentiate w.r.t. x

$$\frac{\arctan(y)}{y^{2}+1}$$