Question

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

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

Answer 1

The correct option is A $e^x\sqrt{\frac{1+x}{1-x}}+c$

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

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

Using $\int e^x(f\left(x\right)+f^{\prime }\left(x\right))dx=e^{x}f\left(x\right)$

And taking $f\left(x\right)=\frac{\sqrt{1-x^2}}{1-x}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{(1-x)\sqrt{1-x^2}}$

$\therefore I=e^x\sqrt{\frac{1+x}{1-x}}+C$