int e^{x}left(frac{1-x}{1+x^{2}}right)^{2} ,d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We know that $\int e^{x}[f(x) + f'(x)] dx = e^{x}f(x) + C$.Given integral is $I = \int e^{x} \left( \frac{1-x}{1+x^{2}} \right)^{2} dx$.Expanding

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We know that $\int e^{x}[f(x) + f'(x)] dx = e^{x}f(x) + C$.Given integral is $I = \int e^{x} \left( \frac{1-x}{1+x^{2}} \right)^{2} dx$.Expanding the numerator: $I = \int e^{x} \frac{1 + x^{2} - 2x}{(1+x^{2})^{2}} dx$.Splitting the fraction: $I = \int e^{x} \left[ \frac{1+x^{2}}{(1+x^{2})^{2}} - \frac{2x}{(1+x^{2})^{2}} \right] dx$.$I = \int e^{x} \left[ \frac{1}{1+x^{2}} - \frac{2x}{(1+x^{2})^{2}} \right] dx$.Let $f(x) = \frac{1}{1+x^{2}}$. Then $f'(x) = -\frac{1}{(1+x^{2})^{2}} \cdot (2x) = -\frac{2x}{(1+x^{2})^{2}}$.Thus,$I = \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.Therefore,$I = \frac{e^{x}}{1+x^{2}} + C$.

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

Similar Questions

$\int_1^2 {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)\,dx = } $

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

$\int e^x(1-\cot x+\cot^2 x) dx =$

$\int {{e^x}(1 + \tan x)\sec x\,dx = }$

Evaluate the definite integral $\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module