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

If $\int e^{2x} \frac{2(\sin 2x \cos 2x - 1)}{2 \sin^2 2x} dx = A e^{2x} \cot 2x + c$ (where $c$ is the constant of integration), then $A^3 =$ ?

If $\int \frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x)+c$, then $f(x)$ is:

$\int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$ is equal to

$\int \frac{e^{\sin x}(\sin 2x - 8 \cos x)}{2(\sin x - 3)^2} dx =$

If an antiderivative of $f(x)$ is $e^x$ and that of $g(x)$ is $\cos x,$ then $\int f(x) \cos x \, dx + \int g(x) e^x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module