Chapter 7.3, Problem 9E

To determine

To find the indefinite integral.

Answer to Problem 9E

  $\text{The indefinite integral is, }{\displaystyle \int y\ln ydy}=\frac{1}{2}{y}^2\ln y-\frac{1}{4}{y}^2+C.$

Explanation of Solution

Given information: $\text{The equation is},{\displaystyle \int y\ln ydy}.$

Concept used: Integration by parts.

  $\begin{array}{l}{\displaystyle \int udv=uv-{\displaystyle \int vdu}}\\ \text{This formula expresses one integral, }{\displaystyle \int udv\text{ , in terms}}\text{ of a second integral, }{\displaystyle \int vdu}.\text{ With a}\\ \text{proper choice of }u\text{ and }v,\text{ the second integral may be easier to evaluate than the first}\text{. This }\\ \text{is the reason for the importance of the formula}\text{. When faced with an integral that we cannot}\\ \text{handle analytically, We can replace it by one with which we might have more success}\text{. }\end{array}$

Calculation : Using integration by parts, indefinite integral is obtained as,

  $\begin{array}{l}\text{For,}\\ {\displaystyle \int udv={\displaystyle \int y\ln ydy}}\left(\text{let }u=\ln y\text{ and }dv=ydy\right)\\ \text{This then gives, }du=\frac{1}{y}dy\text{ and }v=\frac{1}{2}{y}^2.\\ \text{using the integration by parts formula, }{\displaystyle \int udv=uv-{\displaystyle \int vdu,\text{ then gives,}}}\\ {\displaystyle \int y\ln ydy}=\ln y\left(\frac{1}{2}{y}^2\right)-{\displaystyle \int \frac{1}{2}{y}^2}\left(\frac{1}{y}dy\right)\\ =\frac{1}{2}{y}^2\ln y-\frac{1}{2}{\displaystyle \int ydy}\\ =\frac{1}{2}{y}^2\ln y-\frac{1}{2}\left(\frac{1}{2}{y}^2\right)+C\\ =\frac{1}{2}{y}^2\ln y-\frac{1}{4}{y}^2+C\\ \text{Hence, the indefinite integral is, }{\displaystyle \int y\ln ydy}=\frac{1}{2}{y}^2\ln y-\frac{1}{4}{y}^2+C.\end{array}$