#### Provide solution for RD Sharma maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.29 Question 10 maths textbook solution.

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

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

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

Let, $x=a \frac{d}{d x}\left(x^{2}+x\right)+b$

\begin{aligned} &\Rightarrow x=a(2 x+1)+b \\ &\Rightarrow x=2 x a+a+b \end{aligned}

Comparing the coefficient of x and the constant terms, we get

\begin{aligned} &\Rightarrow 2 a=1 \quad \text { and } \quad a+b=0 \\ &\Rightarrow a=1 / 2 \; \; \; \; \;\quad \Rightarrow b=-a \\ &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \Rightarrow b=-\frac{1}{2} \\ &I=\int\left(\frac{1}{2}(2 x+1)+\left(-\frac{1}{2}\right)\right) \sqrt{x^{2}+x} d x \end{aligned}

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

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

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

Use the formula : $\left[\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|\right]+C$

And $\left[\int x^{n} d x=\frac{x^{n}+1}{n+1}+C\right]$

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

\begin{aligned} &I=\frac{1}{3}\left(x^{2}+x\right)^{3 / 2}-\frac{1}{2}\left[\frac{(2 x+1)}{4} \sqrt{x^{2}+x}-\frac{1}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x}\right|\right]+C \\ &I=\frac{1}{3}\left(x^{2}+x\right)^{3 / 2}-\frac{1}{8}(2 x+1) \sqrt{x^{2}+x}+\frac{1}{16} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^{2}+x}\right|+C \end{aligned}