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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 106 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 106 Maths Textbook Solution.

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Answer: I=\frac{1}{3} \log \left|\frac{\sqrt{1+x^{8}}-1}{\sqrt{1+x^{8}+1}}\right|+c

Hint: to solve this equation we have to do with differentiate method.

Given: \int \frac{1}{x \sqrt{1+x^{3}}} d x

Solution:

x^{3}=t

3 x^{2} d x=d t

x^{2} d x=\frac{1}{3} d t

I=\int \frac{x^{2}}{x^{3} \sqrt{1+x^{3}}} d x

I=\frac{1}{3} \int \frac{d t}{t \sqrt{1+t}}

\text { Let } 1+\mathrm{t}=\mathrm{p}^{2} \text { , } \mathrm{dt}=2 \mathrm{pdp}

I=\frac{1}{3} \int \frac{2 p d p}{\left.p^{2}-1\right) p}

I=\frac{2}{3} \int \frac{d p}{\left(p^{2}-1\right)}

\frac{2}{3} \times \frac{1}{3} \times \frac{1}{2} \log \left[\frac{p-1}{p+1}\right]+C \quad\left[\frac{d x}{x^{2}-a^{2}}=\frac{1}{2 a} \log \left[\frac{x-a}{x+a}\right]+C\right]

I=\frac{1}{3} \log \left|\frac{\sqrt{1+x^{3}}-1}{\sqrt{1+x^{3}}+1}\right|+c

 

 

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