Question

${\int }_0^1\frac{dx}{(1+x)\sqrt{(2+x-x^2)}}=\frac{1}{k}\sqrt{2}$. Find the value of $k$.

Answer 1

Let $I=\int \frac{1}{(1+x)\sqrt{2+x-x^2}}=\int \frac{1}{\sqrt{\frac{9}{4}-{(x-\frac{1}{2})}^2}(x+1)}dx$
Substitute $x-\frac{1}{2}=t\Rightarrow dx=dt$
$I=\int \frac{1}{(t+\frac{3}{2})\sqrt{\frac{9}{4}-t^2}}dt$
Substitute $t=\frac{3}{2}\sin u\Rightarrow dt=\frac{3\cos u}{2}dx$
$\therefore I=\int \frac{1}{\frac{3\sin u}{2}+\frac{3}{2}}dt$
Substitute $p=\tan \frac{u}{2}\Rightarrow dp=\frac{1}{2}{\sec }^2\frac{u}{2}du$
$\therefore I=\int \frac{4}{3p^2+6p+3}dp=4\int \frac{1}{{(\sqrt{3}p+\sqrt{3})}^2}dp$
$=-\frac{4}{3p+3}=\frac{2(x-2)}{3\sqrt{-x^2+x+2}}$
$\therefore {\int }_0^1\frac{1}{(1+x)\sqrt{2+x-x^2}}dx=\frac{\sqrt{2}}{3}$