#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18 .19 Question 11 Maths Textbook Solution.

Answer:  $\frac{1}{4} \ln \left(2 x^{2}+6 x+5\right)+\frac{1}{2} \tan ^{-1}(2 x+3)+c$

Hint: Multiply and Divide by 4

Given: $\int \frac{x+2}{2 x^{2}+6 x+5} d x$

Solution:Multiply and divide by 4

$\frac{1}{4} \int \frac{4 x+6}{2 x^{2}+6 x+5} d x+\frac{2}{4} \int \frac{d x}{2 x^{2}+6 x+5}$  .........(1)

\begin{aligned} &I_{1}=\int \frac{4 x+6}{2 x^{2}+6 x+5} d x \\ &2 x^{2}+6 x+5=t \\ &(4 x+6) d x=d t \\ &I_{1}=\int \frac{d t}{t}=\ln t+c=\ln \left(2 x^{2}+6 x+5\right)+c_{1} \end{aligned}

\begin{aligned} &I_{2}=\frac{1}{2} \int \frac{d x}{2 x^{2}+6 x+5} \\ &I_{2}=\frac{1}{2} \int \frac{d x}{x^{2}+3 x+\frac{9}{4}+\frac{5}{2}-\frac{9}{4}} \end{aligned}

$\frac{1}{2} \int \frac{d x}{\left(x+\frac{3}{2}\right)^{2}+\frac{1}{4}}$

\begin{aligned} &\left(x+\frac{3}{2}\right)=u \Rightarrow 1 d x=d u \\ &I_{2}=\frac{1}{2} \int \frac{d u}{u^{2}+\left(\frac{1}{2}\right)^{2}} \\ &\frac{1}{2}\left[\frac{1}{\frac{1}{2}} \tan ^{-1}\left(\frac{u}{\frac{1}{2}}\right)\right]+c_{2}\left[\int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{2}\right. \\ &\begin{array}{l} \tan ^{-1}\left[2 \tan ^{-1}2\left(x+\frac{3}{2}\right)\right]+c_{2} \\ I_{2}=\tan ^{-1}(2 x+3)+c_{2} \\ I=\frac{1}{4} \ln \left(2 x^{2}+6 x+5\right)+c_{1}+\frac{1}{2} \tan ^{-1}(2 x+3)+c_{2} \\ I=\frac{1}{4} \ln \left(2 x^{2}+6 x+5\right)+\frac{1}{2} \tan ^{-1}(2 x+3)+c \end{array} \end{aligned}