Answer
$\pi i+\dfrac{\pi \sqrt 3}{4}k$
Work Step by Step
We need to integrate the given integral as follows:
$[\int_{0}^{1} \dfrac{2}{\sqrt {1-t^2}} dt)]i+[\int_{0}^{1} \dfrac{\sqrt 3}{1+t^2} dt)]k=[2 \sin^{-1} t]_{0}^{1} i+[\sqrt 3 \tan^{-1} t]_{0}^{1} k $
or,
$[2 \sin^{-1} t]_{0}^{1} i+[\sqrt 3 \tan^{-1} t]_{0}^{1} k =(\sin^{-1} (1)-\sin^{-1} (0))i+(\tan^{-1} (1)-\tan^{-1} (0))k$
Thus,
$(\sin^{-1} (1)-\sin^{-1} (0))+(\tan^{-1} (1)-\tan^{-1} (0))=\pi i+\dfrac{\pi \sqrt 3}{4}k$
