#### Provide solution for RD Sharma maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.29 Question 7 maths textbook solution.

Answer : $\inline \\I=\frac{\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\left[\frac{2 x+1}{4} \sqrt{x^{2}+x+1}+\frac{3}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|\right]+C$

Hint : To solve the given integration, we express the linear term as a derivative of quadratic into constant plus another constant

Given : $\inline \int(x+1) \sqrt{x^{2}+x+1} d x$

Solution :

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

For thr first integral: Let, $x^{2}+x+1=t^{2} \Rightarrow(2 x+1) d x=2 t d t$

\begin{aligned} &I=\frac{1}{2} \int 2 t d t+\frac{1}{2} \int \sqrt{x^{2}+2(x)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2}+1-\frac{1}{4}} d x \\ &I=\int t d t+\frac{1}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}} d x \end{aligned}

$I=\frac{t^{1+1}}{1+1}+\frac{1}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]$

\begin{aligned} &I=\frac{t^{2}}{2}+\frac{1}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x \\ &\left(\text { Use the formula: } \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C\right) \end{aligned}

\begin{aligned} &I=\frac{\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\left[\frac{\left(x+\frac{1}{2}\right)}{2} \sqrt{x^{2}+x+1}+\frac{\left(\frac{\sqrt{3}}{2}\right)^{2}}{2} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|\right]+C \\ &I=\frac{\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\left[\frac{2 x+1}{4} \sqrt{x^{2}+x+1}+\frac{3}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|\right]+C \end{aligned}