Q 23. - Calculus

Question

Each expression that follows is the result of a calculation that uses integration by parts. That is, each is an expression of the form $uv - \int v du$ for some functions u and v. Identify u, v, du, and dv, and determine the original integral  $\int u dv$.

$x{\tan }^{-1}x-\int \frac{x}{x^2+1} dx$

Step-by-Step Solution

Answer

$\begin{array}{rcl}u& =& {\tan }^{-1}x\\ v& =& x\\ du& =& \frac{dx}{x^2+1}\\ dv& =& dx\\ \int u dv& =& \int {\tan }^{-1}x dx\end{array}$

Step 1. Given information

Integral is $x{\tan }^{-1}x-\int \frac{x}{x^2+1} dx$

Step 2. Explanation

$\begin{array}{rcl}u& =& {\tan }^{-1}x\\ du& =& \frac{dx}{x^2+1}\\ v& =& x\\ dv& =& dx\\ \int u dv& =& \int {\tan }^{-1}x dx\end{array}$Consider the integral $x{\tan }^{-1}x-\int \frac{x}{x^2+1} dx$