e^x ((1-x)/(1+x^2))^2 dx = Option 1: e^x/(1+x^2) + C Option 2: -e^x/(1+x^2) + C Option 3: e^x/((1+x^2)^2) + C Option 4: -e^x/((1+x^2)^2) + C

Answer: e^x/(1+x^2) + C. Solution: Let f(x)=1/(1+x^2), then f'(x)=(-2x)/((1+x^2)^2). f(x)+f'(x)=((1+x^2)-2x)/((1+x^2)^2)=((1-x)^2)/((1+x^2)^2). Using e^x[f(x)+f'(x)]dx=e^x f(x)+C, the answer is e^x/(1+x^2)+C.

Calculus

Integrals

Medium

common

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

$\frac{e^x}{1+x^2} + C$

$-\frac{e^x}{1+x^2} + C$

$\frac{e^x}{(1+x^2)^2} + C$

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

✅ Correct Option: 1

Let $f(x)=\frac{1}{1+x^2}$ , then $f'(x)=\frac{-2x}{(1+x^2)^2}$ .

$f(x)+f'(x)=\frac{(1+x^2)-2x}{(1+x^2)^2}=\frac{(1-x)^2}{(1+x^2)^2}$ .

Using $\int e^x[f(x)+f'(x)]dx=e^x f(x)+C$ , the answer is $\frac{e^x}{1+x^2}+C$ .

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