int {e^x frac{x^2 + 1}{(x + 1)^2} dx} = | 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 $I = \int e^x \frac{x^2 + 1}{(x + 1)^2} dx$. Rewrite the numerator: $x

Vedclass · Accepted Answer

$\left( \frac{x - 1}{x + 1} \right) e^x + c$

Vedclass · Answer

(A) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Given integral is $I = \int e^x \frac{x^2 + 1}{(x + 1)^2} dx$. Rewrite the numerator: $x^2 + 1 = (x^2 - 1) + 2 = (x - 1)(x + 1) + 2$. So,$I = \int e^x \left[ \frac{(x - 1)(x + 1)}{(x + 1)^2} + \frac{2}{(x + 1)^2} \right] dx$. $I = \int e^x \left[ \frac{x - 1}{x + 1} + \frac{2}{(x + 1)^2} \right] dx$. Let $f(x) = \frac{x - 1}{x + 1}$. Then $f'(x) = \frac{(x + 1)(1) - (x - 1)(1)}{(x + 1)^2} = \frac{x + 1 - x + 1}{(x + 1)^2} = \frac{2}{(x + 1)^2}$. Since the integral is in the form $\int e^x [f(x) + f'(x)] dx$,the result is $e^x f(x) + c$. Thus,$I = e^x \left( \frac{x - 1}{x + 1} \right) + c$.

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

Similar Questions

$\int {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)\,dx} $ is equal to

If $\int {{e^{{x^2}}}\left( {2 - \frac{1}{{{x^2}}}} \right)dx = {e^{{x^2}}}f(x) + C} $ and $f\left( {\frac{1}{2}} \right) = 2$,then $f(1)$ is equal to (where $C$ is an arbitrary constant).

$\int_0^1 \frac{x e^x}{(2+x)^3} d x$ is equal to

$\int {{e^x}\sin x(\sin x + 2\cos x)} \,dx = $

The value of $\int e^{x} \left[ \frac{1+\sin x}{1+\cos x} \right] dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module