Problem - 4621
Compute $$\int\arctan{x}dx$$
$$\int\arctan{x}dx=x\arctan{x}-\int x d(\arctan{x})=x\arctan{x} - \int\frac{x}{1+x^2}dx=x\arctan{x}-\frac{1}{2}\int\frac{d(1+x^2)}{1+x^2}=\boxed{x\arctan{x}-\frac{1}{2}\ln(1+x^2) + C}$$
Compute $$\int\arctan{x}dx$$
$$\int\arctan{x}dx=x\arctan{x}-\int x d(\arctan{x})=x\arctan{x} - \int\frac{x}{1+x^2}dx=x\arctan{x}-\frac{1}{2}\int\frac{d(1+x^2)}{1+x^2}=\boxed{x\arctan{x}-\frac{1}{2}\ln(1+x^2) + C}$$