Question

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

Answer 1

Consider the given integral.

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

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

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

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

Therefore,

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

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

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

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

$I=-\int \frac{dt}{\sqrt{4t-1}}$

Let $p=4t-1$

$\frac{dp}{dt}=4$

$\frac{dp}{4}=dt$

Therefore,

$I=-\frac{1}{4}\int \frac{dp}{\sqrt{p}}$

$I=-\frac{1}{4}\left(2\sqrt{p}\right)+C$

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

On putting the value of $p$, we get

$I=-\frac{1}{2}\left(\sqrt{4t-1}\right)+C$

On putting the value of $t$, we get

$I=-\frac{1}{2}\left(\sqrt{4\left(\frac{1}{1+x}\right)-1}\right)+C$

$I=-\frac{1}{2}\left(\sqrt{\frac{4-1-x}{1+x}}\right)+C$

$I=-\frac{1}{2}\left(\sqrt{\frac{3-x}{1+x}}\right)+C$

Hence, this is the answer.