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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 101 Maths Textbook Solution.

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Answer: \frac{1}{n} \log \left|\frac{\sqrt{1+x^{n}}-1}{\sqrt{1+x^{n}}+1}\right|+C


Given:\int \frac{1}{x \sqrt{1+x^{n}}} d x


I=\int \frac{d x}{x \sqrt{1+x^{n}}}

=\int \frac{x^{n-1} d x}{x^{n-1} x^{1} \sqrt{1+x^{n}}}

=\int \frac{x^{n-1} d x}{x^{n} \sqrt{1+x^{n}}}

\text { putting } x^{n}=t

\Rightarrow n x^{n-1} d x=d t

\Rightarrow x^{n-1} d x=\frac{d t}{n}

I=\frac{1}{n} \int \frac{d t}{t \sqrt{1+t}}

\operatorname{let} 1+t=p^{2}

d t=2 p d p


I=\frac{1}{n} \int \frac{2 p d p}{\left(p^{2}-1\right) p}

=\frac{2}{n} \int \frac{d p}{p^{2}-1^{2}}

=\frac{2}{n} \times \frac{1}{2} \log \left|\frac{p-1}{p+1}\right|+C

=\frac{1}{n} \log \left|\frac{\sqrt{1+t}-1}{\sqrt{1+t}+1}\right|+C

=\frac{1}{n} \log \left|\frac{\sqrt{1+x^{n}}-1}{\sqrt{1+x^{n}}+1}\right|+C



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