Question

${\int }_{-1}^{1/2}\frac{e^x(2-x^2)dx}{(1-x)\sqrt{1-x^2}}$ 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 $\sqrt{3e}$

$\to I={\int }_{-1}^{1/2}\frac{e^x(1+1-x^2)}{(1-x)\sqrt{1-x^2}}dx$

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

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

As $\frac{d}{dx}\left(\frac{\sqrt{1-x^2}}{1-x}\right)=\frac{\frac{(-2x)(1-x)}{2\sqrt{1-x^2}}-\sqrt{1-x^2(-1)}}{(1-x{)}^2}$

$=\frac{-x(1-x)+(1-x^2)}{(1-x{)}^2\sqrt{1-x^2}}$

$=\frac{1-x}{(1-x{)}^2\sqrt{1-x^2}}=\frac{1}{(1-x)\sqrt{1-x^2}}$

Therefore we have $\int e^x\left(f\right(x)+f^{\prime }(x)dx$

which is equal to $e^{x}f\left(x\right){\int }_{-1}^{1/2}$

$=\sqrt{e}f\left(\frac{1}{2}\right)-\frac{1}{e}f(-1)$

$f\left(\frac{1}{2}\right)=\frac{\sqrt{1-\frac{1}{4}}}{1-\frac{1}{2}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$

$f(-1)=0$

$\Rightarrow I=\sqrt{3e}\Rightarrow \left(c\right)$