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#### Please solve RD Sharma maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.29 Question 1 maths textbook solution.

Answer : $-\frac{1}{3}\left(x^{2}-x+1\right)^{\frac{3}{2}}+\frac{3}{8}\left(2 x-1 \sqrt{x^{2}-x+1}+3 \ln \left|x+\frac{1}{2}+\sqrt{x^{2}-x+1}\right|+C\right.$

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

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

Solution :

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

Comparing the coefficient of x and constant terms, we get

\begin{aligned} &\Rightarrow 2 a=1 \Rightarrow a=\frac{1}{2} \\ &\Rightarrow b-a=1 \Rightarrow b=1+a \Rightarrow b=1+\frac{1}{2} \Rightarrow b=\frac{3}{2} \end{aligned}

\begin{aligned} &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 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}-x+1} d x \\ &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-\left(\frac{1}{2}\right)^{2}} d x \end{aligned}

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

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

Usinf formula, $\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$

$\\I=\frac{1}{3} t^{3 / 2}+\frac{3}{2}\left(\frac{x-\frac{1}{2}}{2}\left(\sqrt{\left(\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}\right)}\right)+\frac{\left(\frac{\sqrt{3}}{2}\right)^{2}}{2} \log \left|x-\frac{1}{2}+\sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\right|\right)+C$

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

$\\I=\frac{1}{3}\left(x^{2}-x+1\right)^{3 / 2}+\frac{3}{8}(2 x-1) \sqrt{x^{2}-x+1}+\frac{9}{16} \log \left[x-\frac{1}{2}+\sqrt{x^{2}-x+1}\right]+c$