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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 97 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 97 Maths Textbook Solution.

Answers (1)

Answer:

x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\sqrt{x^{2}-a^{2}}+C\left[\because t=x^{2}+a^{2}\right]

Hint

\text { Put } x=a \tan \theta

Given:

\int \log \left(x+\sqrt{x^{2}+a^{2}}\right) d x

Solution:

I=\int 1_{I I} \cdot \log \left(x+\sqrt{x_{I}^{2}+a^{2}}\right) d x

=\log \left(x+\sqrt{x^{2}+a^{2}}\right) \int 1 d x-\int\left[\frac{d}{d x}\left\{\log \left(x+\sqrt{x^{2}+a^{2}}\right)\right\} \int 1 d x\right]

=\log \left(x+\sqrt{x^{2}+a^{2}}\right) \cdot x-\int\left(\frac{1}{x+\sqrt{x^{2}+a^{2}}}\right) \times\left(1+\frac{1 \times 2 x}{2 \sqrt{x^{2}+a^{2}}}\right) \cdot x \cdot d x

=\log \left(x+\sqrt{x^{2}+a^{2}}\right) \cdot x-\int \frac{x}{\sqrt{x^{2}+a^{2}}} d x

\text { putting } x^{2}+a^{2}=\text { tinthe sec o nd inte gral }

\Rightarrow 2 x d x=d t

\Rightarrow x d x=\frac{d t}{2}

I=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\frac{1}{2} \int \frac{1}{\sqrt{t}} d t

=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\frac{1}{2} \int t^{-\frac{1}{2}} d t

=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\frac{1}{2}\left[\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]+C

=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\sqrt{t}+C

=x \cdot \log \left(x+\sqrt{x^{2}+a^{2}}\right)-\sqrt{x^{2}-a^{2}}+C\left[\because t=x^{2}+a^{2}\right]

 

 

Posted by

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