#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 16 maths Textbook Solution.

Answer: $2 \sqrt{x^{2}+4 x+5}-\log \left|x+2+\sqrt{x^{2}+4 x+5}\right|+c$

Given: $\int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x$

Hint: Simplify the given function

Solution:

\begin{aligned} &I=\int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x \\ &I=\int \frac{2 x+4-1}{\sqrt{x^{2}+4 x+5}} d x \\ &I=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+5}} d x-\int \frac{1}{\sqrt{x^{2}+4 x+4-4+5}} d x \\ &I=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+5}} d x-\int \frac{d x}{\sqrt{(x+2)^{2}+1}} \\ &I=\int \frac{2 x+4}{\sqrt{x^{2}+4 x+5}} d x-\log \left|x+2+\sqrt{x^{2}+4 x+5}\right|+c \end{aligned}

Using....................$\left[\int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c\right]$

Let

\begin{aligned} &x^{2}+4 x+5=y \\ &(2 x+4) d x=d y \end{aligned}

$\Rightarrow \int \frac{2 x+4}{\sqrt{x^{2}+4 x+5}} d x=\int \frac{d y}{\sqrt{y}}=\frac{\sqrt{y}}{\frac{1}{2}}+c$

\begin{aligned} &=2 \sqrt{y}+c \\ &=2 \sqrt{x^{2}+4 x+5}+c \end{aligned}

$I=2 \sqrt{x^{2}+4 x+5}-\log \left|x+2+\sqrt{x^{2}+4 x+5}\right|+c$