Chapter 7, Problem 20RE

To determine

To evaluate the integral.

Answer to Problem 20RE

The required solution is, ${\displaystyle \int x^4\ln xdx}\text{ =}\frac{1}{5}x\ln x-\frac{1}{25}{x}^5+C$ .

Explanation of Solution

Given information: The equation is, ${\displaystyle \int x^4\ln xdx}$ .

Concept used: Indefinite integral.

The family of all antiderivatives of a function $f\left(x\right)$ is a indefinite integral of $f$

With respect to $x$ and is denoted by ${\displaystyle \int f\left(x\right).}$

If F is an function such that $F\text{'}\left(x\right)=f\left(x\right)$ , then ${\displaystyle \int f\left(x\right)}dx=F\left(x\right)+C,$

Where C is an arbitrary constant, called the constant of integration.

Calculation : The required solution is obtained as,

  ${\displaystyle \int x^4\ln xdx}$

Use the integration by parts rule

  $\begin{array}{l}{\displaystyle \int udv}=uv-{\displaystyle \int vdu}\\ \text{ let }u=\ln x,\text{ then }du=\frac{1}{x}dx\\ dv=x^{4}dx,\text{ then }v=\frac{1}{5}{x}^5\\ \text{ =}\ln x.\frac{1}{5}{x}^5-{\displaystyle \int \frac{1}{5}{x}^5.\frac{1}{x}dx}\\ \text{ =}\frac{1}{5}x\ln x-\frac{1}{5}{\displaystyle \int x^{4}dx}\\ \text{ =}\frac{1}{5}x\ln x-\frac{1}{5}.\frac{x^5}{5}+C\\ \text{ =}\frac{1}{5}x\ln x-\frac{1}{25}{x}^5+C\end{array}$

Hence, the required solution is, ${\displaystyle \int x^4\ln xdx}\text{ =}\frac{1}{5}x\ln x-\frac{1}{25}{x}^5+C$ .