Indefinite Integrals — Question 10

Question 10

Find an antiderivative of $$\int \frac{x\,\ln(1+x^2)}{1+x^2}\,dx.$$

Question 10 - Solution

Use the substitution $$u=\ln(1+x^2).$$ Then $$du=\frac{2x}{1+x^2}\,dx, \qquad \frac{1}{2}du=\frac{x}{1+x^2}\,dx.$$

Rewrite the integral: $$\int \frac{x\,\ln(1+x^2)}{1+x^2}\,dx = \frac{1}{2}\int u\,du.$$

Integrate: $$\frac{1}{2}\int u\,du = \frac{1}{4}u^2.$$

Substitute back: $$\boxed{ \frac{1}{4}\bigl(\ln(1+x^2)\bigr)^2 + C }$$