Q 21. - 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  u dv.

$-\frac{\ln x}{x^2}+\int \frac{1}{x^3} dx$

Step-by-Step Solution

Answer

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

Step 1. Given information

Integral is $-\frac{\ln x}{x^2}+\int \frac{1}{x^3} dx$

Step 2. Explanation

Consider the integral $-\frac{\ln x}{x^2}+\int \frac{1}{x^3} dx$

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