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  • Provide solution for RD Sharma maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.29 Question 8 maths textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.29 Question 8 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+3) \sqrt{x^{2}+4 x+3} d x

Solution :

Let, x^{2}+4 x+3=u^{2}

  \begin{aligned} &\quad \Rightarrow(2 x+4) d x=2 u d u \\ &I=\int(2 x+3) \sqrt{x^{2}+4 x+3} d x \\ &I=\int(2 x+4-1) \sqrt{x^{2}+4 x+3} d x \end{aligned}

\begin{aligned} &I=\int(2 x+4) \sqrt{x^{2}+4 x+3} d x-\int \sqrt{x^{2}+4 x+3} \mathrm{dx} \\ &I=\int u(2 u) \mathrm{du}-\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=2 \frac{u^{3}}{3}-\left[\frac{x+2}{2} \sqrt{(x+2)^{2}-1}-\frac{1}{2} \log \left|(x+2)+\sqrt{\left(x^{2}+2\right)^{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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