Q.If $f(x)=\frac{\sin^{-1}x}{\sqrt{1-x^2}}$, $g(x)=e^{\sin^{-1}x}$, then $\int f(x)g(x)dx=$
(A) $e^{\sin^{-1}x}(\sin^{-1}x-1)+c$ (B) $e^{\sin^{-1}x}(1-\sin^{-1}x)+c$ (C) $e^{\sin^{-1}x}(\sin^{-1}x+1)+c$ (D) $e^{\sin^{-1}x}(\sin^{-1}x-1)+c$

The product is $f(x)g(x)=\sin^{-1}x\cdot e^{\sin^{-1}x}\cdot\frac{1}{\sqrt{1-x^2}}$. Let $t=\sin^{-1}x$, so $dt=\frac{dx}{\sqrt{1-x^2}}$. The integral becomes $\int t\,e^t\,dt$. By parts with $u=t$, $dv=e^t\,dt$: $=te^t-\int e^t\,dt=te^t-e^t+c=e^t(t-1)+c$. Substituting back $t=\sin^{-1}x$: $e^{\sin^{-1}x}(\sin^{-1}x-1)+c$. Note the source paper prints options (A …

Missing that $g'(x)$ relates directly to $f(x)$'s denominator, and attempting by-parts on the original $x$-express …