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#### Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.15 Question 4

$\frac{1}{3} \log \left|\frac{4 x-4}{4 x+2}\right|+C$

Hint:

To solve this problem use special integration formula

Given:

$\int \frac{1}{2 x^{2}-x-1} d x$

Solution:

Let    $I=\int \frac{1}{2 x^{2}-x-1} d x=\frac{1}{2} \int \frac{1}{x^{2}-\frac{x}{2}-\frac{1}{2}} d x$

\begin{aligned} &=\frac{1}{2} \int \frac{1}{x^{2}-2 \cdot x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}-\frac{1}{2}} d x \\ & \end{aligned}

$=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\frac{1}{16}-\frac{1}{2}} d x$

\begin{aligned} &=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\left(\frac{1+8}{16}\right)} d x \\ & \end{aligned}

$=\frac{1}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^{2}-\frac{9}{16}} d x$

Put   $x-\frac{1}{4}=t \Rightarrow d x=d t$

Then   $I=\frac{1}{2} \int \frac{1}{t^{2}-\left(\frac{3}{4}\right)^{2}} d x$

$=\frac{1}{2} \cdot \frac{1}{2 \times \frac{3}{4}} \log \left|\frac{t-\frac{3}{4}}{t+\frac{3}{4}}\right|+C$                                 $\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]$

$=\frac{1}{3} \log \left|\frac{x-\frac{1}{4}-\frac{3}{4}}{x-\frac{1}{4}+\frac{3}{4}}\right|+C \text { }$                                                 $\quad\left[\because t=x-\frac{1}{4}\right]$

\begin{aligned} &=\frac{1}{3} \log \left|\frac{\frac{4 x-1-3}{4}}{\frac{4 x-1+3}{4}}\right|+C \\ & \end{aligned}

$=\frac{1}{3} \log \left|\frac{4 x-4}{4 x+2}\right|+C$