Question

If $\int \frac{e^x(2-x^2)}{(1-x)\sqrt{1-x^2}}dx=\mu e^x{\left(\frac{1+x}{1-x}\right)}^{\lambda }+C$, then $2(\lambda +\mu )$ is equal to

• A. $-1$
• B. $0$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

Given $\int \frac{e^x(2-x^2)}{(1-x)\sqrt{1-x^2}}dx=\mu e^x{\left(\frac{1+x}{1-x}\right)}^{\lambda }+C$ .....(1)
Consider $I=\int \frac{e^x(2-x^2)}{(1-x)\sqrt{1-x^2}}dx$
$=\int \frac{e^x(1+1-x^2)}{(1-x)\sqrt{1-x^2}}dx$
$=\int \frac{e^x}{(1-x)\sqrt{1-x^2}}dx+\int \frac{e^x(1-x^2)}{(1-x)\sqrt{1-x^2}}dx$
$=\int \frac{e^x}{(1-x)\sqrt{1-x^2}}dx+\int e^x\sqrt{\frac{1+x}{1-x}}dx$
Now, $\frac{d}{dx}\left(\sqrt{\frac{1+x}{1-x}}\right)=\frac{1}{(1-x)\sqrt{1-x^2}}$
So, our integral is of the form $\int e^x\left(f^{\prime }\right(x)+f(x\left)\right)dx=e^{x}f\left(x\right)+C$
Hence, $I=e^x\left(\sqrt{\frac{1+x}{1-x}}\right)+C$
So, equation (1) becomes
$\therefore I=e^x\sqrt{\frac{1+x}{1-x}}+C$
$I=e^x{\left(\frac{1+x}{1-x}\right)}^{\frac{1}{2}}+C$
On comparing , we get
$\mu =1,\lambda =\frac{1}{2}$
$2(\mu +\lambda )=2(1+\frac{1}{2})=2\times \frac{3}{2}=3$