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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 13 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 13 Maths Textbook Solution.

Answers (1)

Answer: -3 \sqrt{5-x^{2}-2 x}-2 \sin ^{-1}\left(\frac{x+1}{\sqrt{6}}\right)+c

Given: \int \frac{3 x+1}{\sqrt{5-x^{2}-2 x}} d x

Hint: Simplify the given Function

Solution:

           \begin{aligned} &I=\int \frac{3 x+1}{\sqrt{5-x^{2}-2 x}} d x \\ &I=3 \int \frac{x+\frac{1}{3}}{\sqrt{5-x^{2}-2 x}} d x \end{aligned}

Multiplying and dividing the numerator by -2,

\begin{aligned} I &=\frac{-3}{2} \int \frac{-2 x-\frac{2}{3}}{\sqrt{5-x^{2}-2 x}} d x \\ I &=\frac{-3}{2} \int \frac{-2 x-2+\frac{4}{3}}{\sqrt{5-x^{2}-2 x}} d x \\ I &=\frac{-3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}} d x-2 \int \frac{1}{\sqrt{5-x^{2}-2 x}} d x \\ I &=I_{1}+I_{2} \\ I_{1} &=\frac{-3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}} d x ; I_{2}=-2 \int \frac{1}{\sqrt{5-x^{2}-2 x}} d x \end{aligned}

Now, I_{1}=\frac{-3}{2} \int \frac{-2 x-2}{\sqrt{5-x^{2}-2 x}} d x

Let

\begin{aligned} &5-x^{2}-2 x=y \\ &(-2-2 x) d x=d y \end{aligned}

\Rightarrow I_{1}=\frac{-3}{2} \int \frac{d y}{\sqrt{y}}=\frac{-3}{2}\left(\frac{\sqrt{y}}{\frac{1}{2}}\right)+c

\begin{aligned} &I_{1}=-3 \sqrt{y}+c \\ &I_{1}=-3 \sqrt{5-x^{2}-2 x}+c \end{aligned}

Now, I_{2}=-2 \int \frac{1}{\sqrt{5-x^{2}-2 x}} d x

\begin{aligned} &I_{2}=-2 \int \frac{1}{\sqrt{5-\left(x^{2}+2 x+1\right)+1}} d x \\ &I_{2}=-2 \int \frac{1}{\sqrt{6-(x+1)^{2}}} d x \\ &I_{2}=-2 \sin ^{-1}\left[\frac{x+1}{\sqrt{6}}\right]+c \end{aligned}

Putting I_{1} & I_{2} in equation (1) we get

I=-3 \sqrt{5-x^{2}-2 x}-2 \sin ^{-1}\left(\frac{x+1}{\sqrt{6}}\right)+c



 

Posted by

infoexpert21

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