#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.15 Question 1 Maths Textbook Solution.

$\frac{1}{8} \log \left|\frac{2 x+1}{2 x+5}\right|+C$

Hint:

To solve this problem use special integration formula

Given:

$\int \frac{1}{4 x^{2}+12 x+5} d x$

Solution:

Let       $I=\int \frac{1}{4 x^{2}+12 x+5} d x=\frac{1}{4} \int \frac{1}{x^{2}+\frac{12 x}{4}+\frac{5}{4}} d x$

\begin{aligned} &=\frac{1}{4} \int \frac{1}{x^{2}+3 x+\frac{5}{4}} d x=\frac{1}{4} \int \frac{d x}{x^{2}+2 \cdot x \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+\frac{5}{4}} \\\\ &=\frac{1}{4} \int \frac{1}{\left(x+\frac{3}{2}\right)^{2}-\frac{9}{4}+\frac{5}{4}} d x=\frac{1}{4} \int \frac{1}{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{9-5}{4}\right)} d x \end{aligned}

Put     $x+\frac{3}{2}=t \Rightarrow d x=d t$

Then      $I=\frac{1}{4} \int \frac{1}{t^{2}-1^{2}} d t$

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

$=\frac{1}{4} \times \frac{1}{2} \log \left|\frac{x+\frac{3}{2}-1}{x+\frac{3}{2}+1}\right|+C \text { }$                              $\left[\because t=x+\frac{3}{2}\right]$

\begin{aligned} &=\frac{1}{8} \log \left|\frac{\frac{2 x+3-2}{2}}{\frac{2 x+3+2}{2}}\right|+C \\\\ &\end{aligned}

$=\frac{1}{8} \log \left|\frac{2 x+1}{2 x+5}\right|+C$