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

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}