Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.21 Question 18 Maths Textbook Solution.

Answer:  $\sqrt{x^{2}+2 x+3}+\log \left|x+1+\sqrt{x^{2}+2 x+3}\right|+c$

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

Hint: Simplify the given function

Solution:

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

$I=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x+\log \left|(x+1)+\sqrt{x^{2}+2 x+3}\right|+c\left|\int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \right| x+\sqrt{x^{2}+a^{2}}|+c|$

Let

\begin{aligned} &x^{2}+2 x+3=y \\ &(2 x+2) d x=d y \end{aligned}

$\Rightarrow \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x=\int \frac{d y}{\sqrt{y}}=\frac{\sqrt{y}}{\frac{1}{2}}+c$

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

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