#### Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.15 Question 3

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

Hint:

To solve this problem use special integration formula

Given:

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

Solution:

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

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

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

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

$=-\int \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\frac{5}{4}} d x$

Put    $x-\frac{1}{2}=t \Rightarrow d x=d t$

Then    $I=-\int \frac{1}{t^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}} d t$

$=\frac{-1}{2 \cdot \frac{\sqrt{5}}{2}} \log \left|\frac{t-\frac{\sqrt{5}}{2}}{t+\frac{\sqrt{5}}{2}}\right|+C \text { }$                                 $\quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]$

$=\frac{-1}{\sqrt{5}} \log \left|\frac{x-\frac{1}{2}-\frac{\sqrt{5}}{2}}{x-\frac{1}{2}+\frac{\sqrt{5}}{2}}\right|+C$                                           $\quad\left[\because t=x-\frac{1}{2}\right]$

\begin{aligned} &=\frac{-1}{\sqrt{5}} \log \left|\frac{\frac{2 x-1-\sqrt{5}}{2}}{\frac{2 x-1+\sqrt{5}}{2}}\right|+C \\ & \end{aligned}

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