#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 100  Maths Textbook Solution.

$(\log x)^{2} \times \frac{x^{4}}{4}-\frac{\log x \cdot x^{4}}{8}+\frac{x^{4}}{32}+C$

Hint:

You must know about the log formula of integration.

Given:

$\int x^{3}(\log x)^{2} d x$

Solution:

$\int x^{3}{ }_{11} \cdot\left(\log _{1} x\right)^{2} \cdot d x$

$=\left(\log x^{2}\right) \int x^{3} d x-\int \frac{2 \log x}{x} \times \frac{x^{4}}{4} d x$

$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{1}{2} \int \log _{1} x \cdot x^{3}{ }_{I I} d x$

$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{1}{2}\left[\log x \int x^{3} d x-\int\left\{\frac{d}{d x}(\log x) \int x^{3} d x\right\} d x\right]$

$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{1}{2}\left[\log x \cdot \frac{x^{4}}{4}-\int \frac{1}{x} \times \frac{x^{4}}{4} d x\right]$

$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{1}{2}\left[\log x \cdot \frac{x^{4}}{4}-\frac{1}{4} \int x^{3} d x\right]$

$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{1}{2}\left[\log x \cdot \frac{x^{4}}{4}-\frac{x^{4}}{16}\right]+C$

$=(\log x)^{2} \times \frac{x^{4}}{4}-\frac{\log x \cdot x^{4}}{8}+\frac{x^{4}}{32}+C$