Computing Indefinite Integrals — Question 2

Question 2 - Solution

Use integration by parts. Let $$u=\ln(x^2+1), \qquad dv=x^2\,dx.$$ Then $$du=\frac{2x}{x^2+1}\,dx, \qquad v=\frac{x^3}{3}.$$

Apply integration by parts: $$\int x^2\ln(x^2+1)\,dx = \frac{x^3}{3}\ln(x^2+1) - \frac{2}{3}\int \frac{x^4}{x^2+1}\,dx.$$

Simplify the integrand by division: $$\frac{x^4}{x^2+1} = x^2-1+\frac{1}{x^2+1}.$$

Thus, $$\int \frac{x^4}{x^2+1}\,dx = \int (x^2-1)\,dx + \int \frac{1}{x^2+1}\,dx = \frac{x^3}{3}-x+\arctan x.$$

Substitute back: $$\int x^2\ln(x^2+1)\,dx = \frac{x^3}{3}\ln(x^2+1) - \frac{2}{3}\left(\frac{x^3}{3}-x+\arctan x\right).$$

Final Answer: $$\boxed{ \frac{x^3}{3}\ln(x^2+1) -\frac{2x^3}{9} +\frac{2x}{3} -\frac{2}{3}\arctan x + C }$$