Question

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

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

Answer 1

The correct option is A $\frac{e^x(1+x)}{\sqrt{1-x^2}}+c$

$I=\int e^x\frac{(1-x^2)+1}{(1-x)\sqrt{1-x^2}}dx=\int e^x\{\frac{1-x^2}{(1-x)\sqrt{1-x^2}}+\frac{1}{(1-x)\sqrt{1-x^2}}\}dx=\int e^x\{\frac{1+x}{\sqrt{1-x^2}}+\frac{1}{(1-x)\sqrt{1-x^2}}\}dx=\int e^x\{\frac{\sqrt{1+x}}{\sqrt{1-x}}+\frac{1}{(1-x)\sqrt{1-x^2}}\}dx$
$\underset{f\left(x\right)}{\downarrow }\underset{f^{\prime }\left(x\right)}{\downarrow }$
$\{\because \int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+c\}\therefore I=\frac{e^x(1+x)}{\sqrt{1-x^2}}+c$