Computing Indefinite Integrals — Question 6

Question 6 - Solution

Rewrite the integrand by splitting the rational expression: $$\frac{x^3}{1+x^2} = x - \frac{x}{1+x^2}.$$

Thus, $$\int \frac{x^3\ln x}{1+x^2}\,dx = \int x\ln x\,dx - \int \frac{x\ln x}{1+x^2}\,dx.$$

First integral: Use integration by parts with $$u=\ln x, \qquad dv=x\,dx.$$ Then $$du=\frac{1}{x}\,dx, \qquad v=\frac{x^2}{2}.$$ So $$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4}.$$

Second integral: Use integration by parts with $$u=\ln x, \qquad dv=\frac{x}{1+x^2}\,dx.$$ Then $$du=\frac{1}{x}\,dx, \qquad v=\frac12\ln(1+x^2).$$ Hence, $$\int \frac{x\ln x}{1+x^2}\,dx = \frac12\ln x\,\ln(1+x^2) - \frac12\int \frac{\ln(1+x^2)}{x}\,dx.$$

The remaining integral is non-elementary and is expressed using the dilogarithm function: $$\int \frac{\ln(1+x^2)}{x}\,dx = -\frac12\,\operatorname{Li}_2(-x^2).$$

Therefore, $$\int \frac{x\ln x}{1+x^2}\,dx = \frac12\ln x\,\ln(1+x^2) + \frac14\,\operatorname{Li}_2(-x^2).$$

Final Answer: $$\boxed{ \frac{x^2}{2}\ln x - \frac{x^2}{4} - \frac12\ln x\,\ln(1+x^2) - \frac14\,\operatorname{Li}_2(-x^2) + C }$$