#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 17 Maths Textbook Solution.

Answer: $5 \sqrt{x^{2}+4 x+10}-7 \log \left|x+2+\sqrt{x^{2}+4 x+10}\right|$

Given: $\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x$

Hint: Simplify the given function

Solution:

\begin{aligned} &I=\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x\\ &I=5 \int \frac{x+\frac{3}{5}}{\sqrt{x^{2}+4 x+10}} d x \text { multiplying and dividing by } 2 \text { in numerator, }\\ &I=\frac{5}{2} \int \frac{2 x+\frac{6}{5}}{\sqrt{x^{2}+4 x+10}} d x\\ &I=\frac{5}{2} \int \frac{2 x+4-\frac{14}{5}}{\sqrt{x^{2}+4 x+10}} d x \end{aligned}

\begin{aligned} &I=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+10}} d x \\ &I=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}+4 x+4+6}} d x \\ &I=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{(x+2)^{2}+(\sqrt{6})^{2}}} d x \\ &I=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \log \left|x+2+\sqrt{x^{2}+4 x+10}\right| \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+10=y \\ &(2 x+4) d x=d y \end{aligned}

$\Rightarrow \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{d y}{\sqrt{y}}=\frac{\sqrt{y}}{\frac{1}{2}}+c$

\begin{aligned} &=2 \sqrt{y}+c \\ &=2 \sqrt{x^{2}+4 x+10}+c \end{aligned}

$I=5 \sqrt{x^{2}+4 x+10}-7 \log \left|x+2+\sqrt{x^{2}+4 x+10}\right|$