Mathematics - Indefinite Integration Question with Solution | TestHub

Given,

$f\left(x\right)=\int \frac{dx}{\left(3+4x^2\right)\sqrt{4-3x^2}}$

Put $x=\frac{1}{t}, dx=-\frac{1}{t^2}dt$

So, $f\left(x\right)=\int \frac{-dt}{t^2\left(3+\frac{4}{t^2}\right)\sqrt{4-\frac{3}{t^2}}}$

$\Rightarrow f\left(x\right)=\int \frac{-t dt}{\left(3t^2+4\right)\sqrt{4t^2-3}}$

Now let, $4t^2-3={\lambda }^2\Rightarrow 8t dt=2\lambda d\lambda$

$\Rightarrow f\left(x\right)=-\int \frac{\lambda d\lambda }{4·\left(3\left(\frac{{\lambda }^2+3}{4}\right)+4\right)·\lambda }$

$\Rightarrow f\left(x\right)=-\int \frac{d\lambda }{3{\lambda }^2+9+16}$

$\Rightarrow f\left(x\right)=-\int \frac{d\lambda }{3{\lambda }^2+25}$

$\Rightarrow f\left(x\right)=-\frac{1}{3}\int \frac{d\lambda }{{\lambda }^2+\frac{25}{3}}$

$\Rightarrow f\left(x\right)=-\frac{1}{3}\times \frac{\sqrt{3}}{5}{\tan }^{-1}\left(\frac{\sqrt{3}\lambda }{5}\right)+c$

$\Rightarrow f\left(x\right)=-\frac{\sqrt{3}}{15}{\tan }^{-1}\left(\frac{\sqrt{3}\sqrt{\left(4-3x^2\right)}}{5x}\right)+c$

Now using, $f\left(0\right)=0\Rightarrow c=+\frac{\sqrt{3}\pi }{30}$

Hence, $f\left(1\right)=\frac{-\sqrt{3}}{15}{\tan }^{-1}\left(\frac{\sqrt{3}}{5}\right)+\frac{\sqrt{3}}{15}\times \frac{\pi }{2}$

$\Rightarrow f\left(1\right)=\frac{-\sqrt{3}}{15}\left({\tan }^{-1}\left(\frac{\sqrt{3}}{5}\right)-\frac{\pi }{2}\right)$

$\Rightarrow f\left(1\right)=\frac{\sqrt{3}}{15}\left(\frac{\pi }{2}-{\tan }^{-1}\left(\frac{\sqrt{3}}{5}\right)\right)$

$\Rightarrow f\left(1\right)=\frac{\sqrt{3}}{15}{\cot }^{-1}\left(\frac{\sqrt{3}}{5}\right)$

$\Rightarrow f\left(1\right)=\frac{\sqrt{3}}{15}{\tan }^{-1}\left(\frac{5}{\sqrt{3}}\right)$

$\Rightarrow f\left(1\right)=\frac{1}{5\sqrt{3}}{\tan }^{-1}\left(\frac{5}{\sqrt{3}}\right)$

Now comparing with given value of $f\left(1\right)$ we get, $\alpha =5 \& \beta =\sqrt{3}$

$\Rightarrow {\alpha }^2+{\beta }^2=28$