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

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

Answers (1)

Answer:-

The answer is
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

Posted by

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