Question

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

• A. + C
• B. + C
• C. + C
• D.

Answer 1

The correct option is B

+ C

Whenever there is $e^x$ in the integrand the first thing we have to see is that whether it is $\int e^x$ [f(x) + f'(x)]dx form or not. If yes, we'll use the corresponding formula, and if not we'll consider other methods. Let's try to simplify the above given integral and see whether it is of

$\int e^x$ [f(x) + f'(x)]dx form

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

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

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

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

$\frac{-2x}{(1+x^2{)}^2}$ could be the derivative of $\frac{1}{(1+x^2)}$. Let's check.

$\frac{d}{dx}\frac{1}{(1+x^2)}=\frac{(1+x^2)\left(0\right)-2x\left(1\right)}{(1+x^2{)}^2}=\frac{-2x}{(1+x^2{)}^2}$

So, $\int e^x\frac{1}{(1+x^2)}-\frac{2x}{(1+x^2{)}^2}$dx is the form of $\int e^x$ [f(x) + f'(x)]dx and hence the answer for this integral would be equal to $e^x$$\frac{1}{1+x^2}$ + C