#### Explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.28 question 14

$I=\frac{\log x}{2} \sqrt{(\log x)^{2}+16}+8 \log \left|\log x+\sqrt{(\log x)^{2}+16}\right|+c$

Hints:-

\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c \\\\ &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \end{aligned}

Given:-

$I=\int \frac{\sqrt{(\log x)^{2}+16}}{x} d x$

Solution:-

Let,
$I=\int \frac{1}{x} \sqrt{16+(\log x)^{2}} d x$

Let,  $\log x=t$

Differentiating both sides,

$\Rightarrow \frac{1}{x} d x=d t$

Substituting logx with t, we have

\begin{aligned} &I=\int \sqrt{t^{2}+16} d t \\\\ &I=\int \sqrt{t^{2}+4^{2}} d t \end{aligned}

As I match with the form

\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &I=\left\{\frac{t}{2} \sqrt{t^{2}+16}+\frac{16}{2} \log \left|t+\sqrt{t^{2}+16}\right|\right\}+c \end{aligned}

Putting the value of t back.

$I=\frac{\log x}{2} \sqrt{(\log x)^{2}+16}+8 \log \left|\log x+\sqrt{(\log x)^{2}+16}\right|+c$