Question

$\int e^x\left(\frac{1+\sqrt{1-x^2}{\sin }^{-1}x}{\sqrt{1-x^2}}\right)dx=$

• A. $\frac{e^x}{\sqrt{1-x^2}}+c$
• B. $e^x{\sin }^{-1}x+c$
• C. $e^x(e^{si{n}^{-1}x}+\frac{1}{\sqrt{1-x^2}})+c$
• D. $e^{si{n}^{-1}x}+\frac{1}{\sqrt{1-x^2}}+c$

Answer 1

Answer: (B)

$e^x{\sin }^{-1}x+c$

$\int e^x(\frac{1}{\sqrt{1-x^2}}+{\sin }^{-1}x)dx$
it is in the form of
$\int e^x\left(f\right(x)+f^1(x\left)dx\right)$
where $f\left(x\right)={\sin }^{-1}x$
$=\int e^{x}f\left(x\right)+\int e^x{f}^1\left(x\right)dx-\left(1\right)$

$\int e^x{f}^1\left(x\right)dx=e^x\int f^1\left(x\right)dx-\int e^x(\int f^1(x\left)dx\right)dx$....(using ILATE Rule)

$=e^{x}f\left(x\right)+c-\int e^{x}f\left(x\right)dx-\left(2\right)$

from (1) & (2)
$=e^{x}f\left(x\right)+c$
so,
$\int e^x(\frac{1}{\sqrt{1-x^2}}+{\sin }^{-1}x)=e^x{\sin }^{-1}x+c$