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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 53 Maths Textbook Solution.

Answers (1)

Answer:

I=\left[\sqrt{x^{2}+x}+\ln (\sqrt{x}+\sqrt{x+1})\right]+c

 

ùTo solve this equation we have to take x = u².

Solution: 

I=\int \sqrt{\frac{(x+1)}{x}} \mathrm{dx}

\text { Let } x=u^{2}

d x=2 u d u

I=\int \sqrt{\frac{\left(u^{2}+1\right)}{\mathrm{u}^{2}}} 2 \mathrm{udu}

 

I=\int 2 \sqrt{u^{2}+1} d u                            \left[\because \int \sqrt{x^{2}+a^{2}}=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \ln x+\sqrt{x^{2}+a^{2}}+c\right]

I=2\left[\frac{u}{2} \sqrt{u^{2}+1}+\frac{1}{2} \ln \left(u+\sqrt{u^{2}+1}\right)\right]+c

I=2\left[\frac{\sqrt{x}}{2} \sqrt{x+1}+\frac{1}{2} \ln (\sqrt{x}+\sqrt{x+1})\right]+c

I=\left[\sqrt{x^{2}+x}+\ln (\sqrt{x}+\sqrt{x+1})\right]+c

 

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