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

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

Question

(D) To evaluate $I = \int_0^1 \frac{x e^x}{(2+x)^3} d x$,we rewrite the integrand as follows:$\frac{x}{(2+x)^3} = \frac{(x+2)-2}{(2+x)^3} = \frac{1}{(

Vedclass · Accepted Answer

$\frac{1}{9} \cdot e-\frac{1}{4}$

Vedclass · Answer

(D) To evaluate $I = \int_0^1 \frac{x e^x}{(2+x)^3} d x$,we rewrite the integrand as follows:$\frac{x}{(2+x)^3} = \frac{(x+2)-2}{(2+x)^3} = \frac{1}{(2+x)^2} - \frac{2}{(2+x)^3}$.Let $f(x) = \frac{1}{(2+x)^2}$. Then $f'(x) = -2(2+x)^{-3} = -\frac{2}{(2+x)^3}$.Thus,the integral becomes $\int_0^1 e^x [f(x) + f'(x)] d x$.Using the standard result $\int e^x [f(x) + f'(x)] d x = e^x f(x) + C$,we get:$I = [e^x \cdot \frac{1}{(2+x)^2}]_0^1$.Evaluating at the limits:$I = (e^1 \cdot \frac{1}{(2+1)^2}) - (e^0 \cdot \frac{1}{(2+0)^2}) = \frac{e}{9} - \frac{1}{4}$.

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

Similar Questions

If $\int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx = f(x) + \text{constant}$, then $f(x)$ is equal to

The value of $\int e^{x}(x^{5}+5x^{4}+1)dx$ is

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

Find : $\int \frac{(x^{2}+1) e^{x}}{(x+1)^{2}} d x$

$\int \frac{e^{\cot x}}{\sin^2 x} (2 \log \csc x + \sin 2 x) dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module