The value of I = int frac{(x-1) e^x}{(x+1)^3} | vedclass

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

Question

(D) $I = \int \frac{(x-1) e^x}{(x+1)^3} \,dx$ $I = \int \left( \frac{x+1-2}{(x+1)^3} \right) e^x \,dx$ $I = \int \left[ \frac{x+1}{(x+1)^3} - \frac{2}

Vedclass · Accepted Answer

$\frac{e^x}{(x+1)^2} + C$,(where $C$ is a constant of integration)

Vedclass · Answer

(D) $I = \int \frac{(x-1) e^x}{(x+1)^3} \,dx$ $I = \int \left( \frac{x+1-2}{(x+1)^3} \right) e^x \,dx$ $I = \int \left[ \frac{x+1}{(x+1)^3} - \frac{2}{(x+1)^3} \right] e^x \,dx$ $I = \int \left[ \frac{1}{(x+1)^2} - \frac{2}{(x+1)^3} \right] e^x \,dx$ Let $f(x) = \frac{1}{(x+1)^2}$,then $f'(x) = -2(x+1)^{-3} = \frac{-2}{(x+1)^3}$. Using the formula $\int e^x (f(x) + f'(x)) \,dx = e^x f(x) + C$,we get: $I = e^x \left( \frac{1}{(x+1)^2} \right) + C = \frac{e^x}{(x+1)^2} + C$.

The value of $I = \int \frac{(x-1) e^x}{(x+1)^3} \,dx$ is

Similar Questions

The value of the integral $\int_{0}^{\infty} e^{-2x} (\sin 2x + \cos 2x) dx$ is:

If $\int {\frac{{{e^x}(1 + \sin x)}}{{1 + \cos x}}} dx = {e^x}f(x) + c$,then $f(x) = $

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

$\int e^{\cos ^{-1} x} \left[ \frac{x-\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}} \right] dx =$

$\int {{e^{\sin x}}\left( {\sin x + {{\sec }^2}x} \right)} \,dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module