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#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.32 Question 2

Answer : -     $\frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{2 x+3}-\sqrt{5}}{\sqrt{2 x+3}+\sqrt{5}}\right|+C$

Hint :-            Use substitution method to solve this integral.

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

Sol : -             Let $\mathrm{I}=\int \frac{1}{(x-1) \sqrt{2 x+3}} d x$

Put $2 x+3=t^{2} \Rightarrow 2 d x=2 t d t \rightarrow d x=t dt$ .then,

\begin{aligned} &=\int \frac{1}{\left(\frac{t^{2}-5}{2}\right)} d t \\ &=\int \frac{2}{t^{2}-5} d t \\ &=\int \frac{2}{t^{2}-\left(\sqrt{\left.5^{2}\right)}\right.} d t \end{aligned}

\begin{aligned} &=2 \cdot \frac{1}{2 \sqrt{5}} \log \left|\frac{t-\sqrt{5}}{t+\sqrt{5}}\right|+C \quad\left(\because \int \frac{1}{x^{2}-a^{2}} d x=1 / 2 a \log \left|\frac{x-a}{x+a}\right|+C\right) \\ &=\frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{2 x+3}-\sqrt{5}}{\sqrt{2 x+3}+\sqrt{5}}\right|+C \quad(\because t=\sqrt{2 x+3}) \text { Ans.. } \end{aligned}.