#### Please solve RD Sharma maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.29 Question 4 maths textbook solution.

Answer : $\inline \\\frac{1}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+\frac{3(2 x+1)}{8} \sqrt{x^{2}+x+1}+\frac{9}{16} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|+C$

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

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

Solution :

\begin{aligned} &\text { Let }(x+2)=a \frac{d}{d x}\left(x^{2}+x+1\right)+b \\ &\Rightarrow x+2=a(2 x+1)+b \\ &\Rightarrow x+2=2 a x+a+b \end{aligned}

Now comparing the coefficients of x and the constant term, we get

\begin{aligned} &2 a=1 \Rightarrow a=\frac{1}{2} \text { and } \\ &a+b=2 \Rightarrow b=2-a=2-\frac{1}{2}=\frac{3}{2} \\ &I=\int\left(\frac{1}{2}(2 x+1)+\frac{3}{2}\right) \sqrt{x^{2}+x+1} d x \\ &I=\int \frac{1}{2}(2 x+1) \sqrt{x^{2}+x+1} d x+\int \frac{3}{2} \sqrt{x^{2}+x+1} d x \end{aligned}

For the first integral, let $x^{2}+x+1=t \Rightarrow(2 x+1) d x=d t$

\begin{aligned} &I=\frac{1}{2} \int \sqrt{t} d t+\frac{3}{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=\frac{1}{2} \frac{t^{\frac{1}{2}+1}}{\frac{3}{2}}+\frac{3}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right] \end{aligned}

$I=\frac{t^{\frac{3}{2}}}{3}+\frac{3}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x$

$\left(\text { Use the formula: } \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C\right)$

\begin{aligned} &I=\frac{\left(x^{2}+x+1\right)^{\frac{3}{2}}}{3}+\frac{3}{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{1}{3}\left(x^{2}+x+1\right)^{3 / 2}+\frac{3(2 x+1)}{8} \sqrt{x^{2}+x+1}+\frac{9}{16} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x+1}\right|+C \end{aligned}