#### Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 54  Maths Textbook Solution.

Answer:$\frac{4\sqrt{2}}{3}$

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{0}^{1} \frac{1}{\sqrt{1+x}-\sqrt{x}} d x$

Solution:

Let,$I=\int_{0}^{1} \frac{1}{\sqrt{1+x}-\sqrt{x}} d x$

\begin{aligned} &=\int_{0}^{1}\left(\frac{1}{\sqrt{1+x}-\sqrt{x}} \times \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}\right) d x \\ &=\int_{0}^{1}\left(\frac{\sqrt{1+x}+\sqrt{x}}{(\sqrt{1+x}-\sqrt{x})(\sqrt{1+x}+\sqrt{x})}\right) d x \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right] \\ &=\int_{0}^{1}\left(\frac{\sqrt{1+x}+\sqrt{x}}{(\sqrt{1+x})^{2}-(\sqrt{x})^{2}}\right) d x \\ &=\int_{0}^{1}\left(\frac{\sqrt{1+x}+\sqrt{x}}{1+x-x}\right) d x \\ &=\int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1} d x \\ &=\int_{0}^{1} \sqrt{1+x} d x+\int_{0}^{1} \sqrt{x} d x \end{aligned}

Put $1+x=t\Rightarrow dx=dt$

\begin{aligned} &I=\int_{0}^{1} \sqrt{t} d t+\int_{0}^{1} \sqrt{x} d x \\ &=\int_{0}^{1} t^{\frac{1}{2}} d t+\int_{0}^{1} x^{\frac{1}{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=\left[\frac{t^{\frac{1}{2}}+1}{\frac{1}{2}+1}\right]_{0}^{1}+\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1} \end{aligned}

\begin{aligned} &=\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}+\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1} \\ &=\frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{1}+\frac{2}{3}\left[x^{\frac{3}{2}}\right]_{0}^{1} \\ &=\frac{2}{3}\left[(1+x)^{\frac{3}{2}}\right]_{0}^{1}+\frac{2}{3}\left[x^{\frac{3}{2}}\right]_{0}^{1} \\ &=\frac{2}{3}\left[(1+1)^{\frac{3}{2}}-(1+0)^{\frac{3}{2}}\right]_{0}^{1}+\frac{2}{3}\left[1^{\frac{3}{2}}-0^{\frac{3}{2}}\right]_{0}^{1} \\ &=\frac{2}{3}\left[2^{\frac{3}{2}}-1^{\frac{3}{2}}\right]+\frac{2}{3}\left[1^{\frac{3}{2}}\right] \\ &=\frac{2}{3}\left[\sqrt{2}^{2 \frac{3}{2}}-1+1\right] \end{aligned}

$=\frac{2}{3}\left ( \sqrt{2} \right )^{2}$

$=\frac{2}{3}2\sqrt{2}$

$=\frac{4\sqrt{2}}{3}$