## Calculus (3rd Edition)

Given $$\int x^2\ln(x^2+1)dx$$ Use integration by parts , let \begin{aligned} u&= \ln(x^2+1)\ \ \ \ \ \ \ &dv&= x^2dx\\ du&= \frac{2xdx}{x^2+1}\ \ \ \ \ \ \ & v&= \frac{x^3}{3} \end{aligned} Then \begin{aligned} \int \ln(x^2+1)dx&=\frac{x^3}{3}\ln(x^2+1)-\frac{2}{3}\int \frac{x^4dx}{x^2+1} \end{aligned} To evaluate $\int \dfrac{x^4dx}{x^2+1}$, let $$x=\tan \theta \ \ \ \ \ dx=\sec^2 \theta d\theta$$ Then \begin{aligned} \int\dfrac{x^4dx}{x^2+1}&=\int \dfrac{\tan^4\theta \sec^2\theta d\theta }{\tan^2\theta +1}\\ &=\int \tan^4\theta d\theta \\ &=\int( \sec^2 \theta-1) \tan^2 \theta d\theta \\ &=\int \sec^2 \theta\tan^2\theta d\theta-\int \tan^2 \theta d\theta \\ &=\int \sec^2 \theta\tan^2\theta d\theta-\int (\sec^2 \theta-1) d\theta \\ &=\frac{1}{3}\tan^3 \theta -\tan \theta + \theta +c\\ &= \frac{1}{3}x^3 -x + \tan^{-1}(x)+c \end{aligned} Hence \begin{aligned} \int x^2 \ln(x^2+1)dx&=\frac{x^3}{3}\ln(x^2+1)-\frac{2}{3}\left( \frac{1}{3}x^3 -x + \tan^{-1}(x)+c\right) \end{aligned}