Question

The value of $\int \frac{e^x(2-x^2)dx}{(1-x)\sqrt{1-x^2}}$ 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. $e^x\sqrt{\frac{1-x}{1+x}}+c$
• E. $\sqrt{\frac{1+x}{1-x}}+c$

Answer 1

Answer: (A)

Let $I=\int \frac{e^x(2-x^2)dx}{(1-x)\sqrt{1-x^2}}$ $=\int \frac{e^x(1-x^2)}{(1-x)\sqrt{1-x^2}}dx+\int \frac{e^x}{(1-x)\sqrt{1-x^2}}dx$ $=\int e^x\sqrt{\frac{1+x}{1-x}}dx+\int \frac{e^x}{(1-x)\sqrt{1-x^2}}dx$ $=\left[e^x\sqrt{\frac{1+x}{1-x}}-\frac{1}{2}\int \frac{d}{dx}\sqrt{\frac{1+x}{1-x}}{e}^{x}dx\right]$ $+\int \frac{e^x}{(1-x)\sqrt{1-x^2}}dx$ $=e^x\sqrt{\frac{1+x}{1-x}}-\int \frac{e^x}{(1-x)\sqrt{1-x^2}}dx$ $+\int \frac{e^x}{(1-x)\sqrt{1-x^2}}dx$ $=e^x\sqrt{\frac{1+x}{1-x}}+c$