#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.18 Question 2 Maths Textbook Solution.

Answer: $\frac{1}{2} \log \left|x^{2}+x+3\right|+\frac{1}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)+c$

Hint: Integrate by ILATE

Given: $\int \frac{x+1}{x^{2}+x+3} d x$

Solution:

\begin{aligned} &I_{1}=\frac{1}{2} \int \frac{2 x+2}{x^{2}+x+3} d x \\ &I_{1}=\frac{1}{2} \int \frac{(2 x+1)+1}{x^{2}+x+3} d x \end{aligned}

$=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+3} d x+\frac{1}{2} \int \frac{1}{x^{2}+x+3} d x$

\begin{aligned} I_{1} &=\frac{1}{2} \int \frac{2 x+2}{x^{2}+x+3} d x \\ I_{1} &=\frac{1}{2} \int \frac{(2 x+1)+1}{x^{2}+x+3} d x \\ &=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+3} d x+\frac{1}{2} \int \frac{1}{x^{2}+x+3} d x \end{aligned}                                                                                            $\left[\int \frac{d t}{t}=\ln t+c,\right]$

Let,$x^{2}+x+3=t$

$(2 x+1) d x=d t$

$I_{1}=\log \left|x^{2}+x+3\right|+\mathrm{C}$

$I_{2}=\int \frac{1}{x^{2}+x+3} d x$

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

$=\int \frac{1}{\left(x+\frac{1}{2}\right)^{2}+\frac{11}{4}} d x$                                                        $\left[\begin{array}{c} x+\frac{1}{2}=t \Rightarrow d x=d t \\ \sqrt{\frac{11}{4}}=a=\frac{\sqrt{11}}{2} \end{array}\right]$

$=\frac{1}{\frac{\sqrt{11}}{2}} \tan ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{11}}{2}}\right)+C_{2}$                                    $\left[\int \frac{1}{t^{2}+a^{2}} d t=\frac{1}{a} \tan ^{-1}\left(\frac{t}{a}\right)+C_{2}\right]$

$I_{2}=\frac{2}{\sqrt{11}} \tan ^{-1} \frac{(2 x+1)}{\sqrt{11}}+C_{2}$

$I=\frac{1}{2}\left(I_{1}+I_{2}\right)$

$=\frac{1}{2}\left[\log \left|x^{2}+x+3\right|+\frac{2}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)\right]+c$

$=\frac{1}{2} \log \left|x^{2}+x+3\right|+\frac{1}{\sqrt{11}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{11}}\right)+c$