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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 15 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 15 Maths Textbook Solution.

Answers (1)

Answer: 2 \sqrt{x^{2}+4 x+3}-3 \log \left|x+2+\sqrt{x^{2}+4 x+3}\right|+c

Given: \int \frac{2 x+1}{\sqrt{x^{2}+4 x+3}} d x

Hint: Simplify the given function

Solution:

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

         \begin{aligned} &I=I_{1}-3 \log \left|x+2+\sqrt{x^{2}+4 x+3}\right|+c \quad \ldots\left[\int \frac{1}{\sqrt{x^{2}-a^{2}}} d x=\log \left|x+\sqrt{x^{2}-a^{2}}\right|+c\right] \\ &I_{1}=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+3}} d x \end{aligned}

       Let,

       \begin{aligned} &x^{2}+4 x+3=y \\ &(2 x+4) d x=d y \\ &I_{1}=\int \frac{d y}{\sqrt{y}}=\frac{\sqrt{y}}{\frac{1}{2}}+c \\ &I_{1}=2 \sqrt{x^{2}+4 x+3}+c \\ &I=2 \sqrt{x^{2}+4 x+3}-3 \log \left|x+2+\sqrt{x^{2}+4 x+3}\right|+c \end{aligned}

 

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