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Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 1 maths Textbook Solution.

Answers (1)

Answer: $\sqrt{x^{2}+6 x+10}-3\left[\log \left|(x+3)+\sqrt{x^{2}+6 x+10}\right|\right]+c$

Given: $\int \frac{x}{\sqrt{x^{2}+6 x+10}} d x$

Hint: Simplify it and integrate it with the help of formula.

Solution: Let $I=\int \frac{x}{\sqrt{x^{2}+6 x+10}} d x$

$=\int \frac{x+3-3}{\sqrt{x^{2}+6 x+10}} d x$ ......................Adding +3 and -3 in numerator

$I=\frac{1}{2} \int \frac{2(x+3)}{\sqrt{x^{2}+6 x+10}} d x-3 \int \frac{d x}{\sqrt{x^{2}+6 x+10}}$

$I=I_{1}-I_{2}$ ..................(1)

Where

$\begin{aligned} &I_{1}=\frac{1}{2} \int \frac{2(x+3)}{\sqrt{x^{2}+6 x+10}} d x \\ &I_{2}=3 \int \frac{d x}{\sqrt{x^{2}+6 x+10}} \end{aligned}$

First solve $I_{1}$

Let

$\begin{aligned} &x^{2}+6 x+10=y \\ &(2 x+6) d x=d y \end{aligned}$ .........(2)

$I_{1}=\frac{1}{2} \int \frac{d y}{\sqrt{y}}=\frac{1}{2} \int(y)^{-\frac{1}{2}} d y$

$I_{1}=\frac{1}{2}\left[\frac{y^{\frac{1}{2}}}{\frac{1}{2}}\right]+c$

$I_{1}=\sqrt{y+c}$ From equation (2)

$I_{1}=\sqrt{x^{2}+6 x+10}+c$

Now $I_{2}=3 \int \frac{d x}{\sqrt{x^{2}+6 x+10}}$

$x^{2}+6 x+10=x^{2}+6 x+9-9+10$

$=\left ( x+3 \right )^{2}+1$

$I_{2}=3 \int \frac{d x}{\sqrt{(x+3)^{2}+1}}$

$I_{2}=3 \log \left|x+3+\sqrt{x^{2}+6 x+10}\right|+c$

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

Now, putting value of $I_{1}$ & $I_{2}$ in equation 1

$I=\sqrt{x^{2}+6 x+10}-3\left[\log \left|(x+3)+\sqrt{x^{2}+6 x+10}\right|\right]+c$

Posted by

infoexpert21

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