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

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

Answers (1)

Answer : 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}

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