Question

$\underset{-1}{\overset{\frac{1}{2}}{\int }}\frac{e^x(2-x^2)}{(1-x)\sqrt{1-x^2}}dx$ is equal to

• A. $\frac{\sqrt{e}}{2}(\sqrt{3}+1)$
• B. $\frac{\sqrt{3e}}{2}$
• C. $\sqrt{3e}$
• D. $\sqrt{\frac{e}{3}}$

Answer 1

The correct option is C $\sqrt{3e}$

Let $I=\underset{-1}{\overset{\frac{1}{2}}{\int }}\frac{e^x(2-x^2)}{(1-x)\sqrt{1-x^2}}dx$
$=\underset{-1}{\overset{\frac{1}{2}}{\int }}\frac{e^x(1-x^2+1)dx}{(1-x)\sqrt{1-x^2}}=\underset{-1}{\overset{\frac{1}{2}}{\int }}{e}^x[\sqrt{\frac{1+x}{1-x}}+\frac{1}{(1-x)\sqrt{1-x^2}}]dx={\left[e^x\sqrt{\frac{1+x}{1-x}}\right]}_{-1}^{\frac{1}{2}}\{\because \int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+c\}=\sqrt{3e}-0=\sqrt{3e}$