Question

Integrate
$\int \frac{dx}{\left(x+1\right)\sqrt{2x^2+3x+1}}$

• A. $I=\left(\sqrt{\frac{2x+1}{1+x}}\right)+C$
• B. $I=2\left(\sqrt{\frac{2x+1}{1+x}}\right)+C$
• C. $I=2\left(\sqrt{\frac{2x-1}{1+x}}\right)+C$
• D. None of these

Answer 1

Answer: (B)

$I=2\left(\sqrt{\frac{2x+1}{1+x}}\right)+C$

Consider the given integral.

$I=\int \frac{dx}{(1+x)\sqrt{2x^2+3x+1}}$ ……. (1)

Let $t=\frac{1}{1+x}$

$\frac{dt}{dx}=-\frac{1}{{(1+x)}^2}$

$dx=-{(1+x)}^{2}dt$

Therefore,

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

$I=-\int \frac{dt}{t\sqrt{2{\left(\frac{1-t}{t}\right)}^2+3\left(\frac{1-t}{t}\right)+1}}$

$I=-\int \frac{dt}{t\sqrt{2\left(\frac{1+t^2-2t}{t^2}\right)+3\left(\frac{1-t}{t}\right)+1}}$

$I=-\int \frac{dt}{t\sqrt{\frac{2(1+t^2-2t)+3t(1-t)+t^2}{t^2}}}$

$I=-\int \frac{dt}{\sqrt{2-4t+3t}}$

$I=-\int \frac{dt}{\sqrt{2-t}}$

Let $p=2-t$

$\frac{dp}{dt}=-1$

$-dp=dt$

Therefore,

$I=\int \frac{dp}{\sqrt{p}}$

$I=2\sqrt{p}+C$

On putting the value of $p$, we get

$I=2\left(\sqrt{2-t}\right)+C$

On putting the value of$t$, we get

$I=2\left(\sqrt{2-\left(\frac{1}{1+x}\right)}\right)+C$

$I=2\left(\sqrt{\frac{2+2x-1}{1+x}}\right)+C$

$I=2\left(\sqrt{\frac{2x+1}{1+x}}\right)+C$

Hence, this is the answer.