Question

The value of $\int \frac{e^x(2-x^2)}{(1-x)\sqrt{1-x^2}}dx$ is equal to

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

Answer 1

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

$\int \frac{e^x(2-x^2)dx}{(1-x)\sqrt{1-x^2}}$
$=\int \frac{e^x(1+1-x^2)dx}{(1-x)\sqrt{1-x^2}}$
$=\int \frac{e^x(1+(1+x\left)\right(1-x\left)\right)dx}{(1-x)\sqrt{1-x^2}}$
$=\int e^x(\frac{(1+x)(1-x)}{(1-x)\sqrt{(1+x)(1-x))}}+\frac{1}{(1-x)\sqrt{1-x^2}})$
$=\int e^x(\sqrt{\frac{1+x}{1-x}}+\frac{1}{(1-x)\sqrt{1-x^2}})$
The second term is the derivative of the first term.
So,
$=e^x\sqrt{\frac{1+x}{1-x}}$