#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.3 Question 6 Maths Textbook Solution.

Answer: $\frac{1}{18}\left\{(2 x+3)^{\frac{3}{2}}-(2 x-3)^{\frac{8}{2}}\right\}+c$

Hint:$\text { To solve this equation we multiply } \sqrt{2 x+3}-\sqrt{2 x-3} \text { to numerator and denominator }$

Given: $\int \frac{1}{\sqrt{2 x+3}+\sqrt{2 x-3}} d x$

Solution: $\int \frac{1}{\sqrt{2 x+3}+\sqrt{2 x-3}} d x$

$\int \frac{\sqrt{2 x+3}-\sqrt{2 x-3}}{(\sqrt{2 x+3}+\sqrt{2 x-3})(\sqrt{2 x+3}-\sqrt{2 x-3})} d x$

$\because \text { multiply } \sqrt{2 x+3}-\sqrt{2 x-3} \text { to numerator and denominator }$

\begin{aligned} &=\int \frac{\sqrt{2 x+3}-\sqrt{2 x-3}}{2 x+3-2 x+3} d x \\ &=\frac{1}{6} \int \sqrt{2 x+3} d x-\int \sqrt{2 x-3} d x\left[\int(a x+b)^{n}=\frac{(a x+b)^{n+1}}{a(n+1)}+c, n \neq 1\right] \\ &=\frac{1}{6}\left[\frac{1}{2} \frac{(2 x+3)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{(2 x-3)^{\frac{1}{2}+1}}{2\left(\frac{1}{2}+1\right)}+c\right] \\ &=\frac{1}{18}\left\{(2 x+3)^{\frac{3}{2}}-(2 x-3)^{\frac{8}{2}}\right\}+c \end{aligned}