2006 SMT/Calculus Problems/Problem 5

Solution

Using integration by parts, we find that the desired integral is equal to:

$\begin{align*} \int(x\tan^{-1}x)dx &= \frac{x^2}{2}\tan^{-1}x-\int\frac{x^2}{2(x^2+1)}dx\\ &=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\int\frac{x^2}{x^2+1}dx\\ &=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\int \left(1-\frac{1}{x^2+1}\right)dx\\ &=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\left(\int 1 dx + \int\frac{-1}{x^2+1}dx\right)\\ &=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\left(x+\cot^{-1}x \right)\\\ \end{align*}$

Therefore the answer is $\boxed{\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\left(x+\cot^{-1}x \right)}$

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2006 SMT/Calculus Problems/Problem 5

Solution