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  • Provide solution for RD Sharma maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.29 Question 9 maths textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.29 Question 9 maths textbook solution.

Answers (1)

Answer : \\I=\frac{2}{3}\left(x^{2}-4 x+3\right)^{3 / 2}-\frac{(x-2)}{2} \sqrt{x^{2}-4 x+3}-\frac{1}{2} \log \left|(x-2)+\sqrt{x^{2}-4 x+3}\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(2 x-5) \sqrt{x^{2}-4 x+3} d x

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

For the first integral :

Let x^{2}-4 x+3=t

\begin{aligned} &\Rightarrow 2 x-4=\frac{d t}{d x} \\ &\Rightarrow(2 x-4) d x=d t \end{aligned}

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

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

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