#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 21 Maths Textbook Solution.

$=\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{x^{2}}{4}+c$

Hint: Taking $(\log x)^{2}$as first function and 1-> Second function and integration

$\int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x$

Given:

$\int x(\log x)^{2} d x$

Solution:

$I=\int x(\log x)^{2} d x$

Taking $(\log x)^{2}$as first function and x  as Second function and integrate

\begin{aligned} &I=(\log x)^{2} \int x d x-\int\left\{\left[\frac{d}{d x}(\log x)^{2} \int x d x\right]\right\} d x \\ &=\frac{x^{2}}{2}(\log x)^{2}-\left[\int 2 \log x \frac{1}{x} \frac{x^{2}}{2} d x\right] \Rightarrow \frac{x^{2}}{2}(\log x)^{2}-\int x \log x d x \end{aligned}

Again integrating by parts, we obtain

\begin{aligned} &I=\frac{x^{2}}{2}(\log x)^{2}-\left[\log x \int x d x-\int\left[\frac{d}{d x} \log x\right] \int x d x\right] d x \\ &=\frac{x^{2}}{2}(\log x)^{2}-\left[\frac{x^{2}}{2} \log x+\int \frac{1}{x} \frac{x^{2}}{2} d x\right] \\ &=\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{1}{2} \int x d x \\ &=\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{x^{2}}{4}+c \end{aligned}