Evaluate the integral: int frac{x e^{2x}}{(1+ | vedclass

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

Question

(B) We are given the integral $I = \int \frac{x e^{2x}}{(1+2x)^2} dx$. Let $2x = t$,then $2 dx = dt$,which implies $dx = \frac{dt}{2}$. Substituting t

Vedclass · Accepted Answer

$\frac{e^{2x}}{4(1+2x)} + C$

Vedclass · Answer

(B) We are given the integral $I = \int \frac{x e^{2x}}{(1+2x)^2} dx$. Let $2x = t$,then $2 dx = dt$,which implies $dx = \frac{dt}{2}$. Substituting these into the integral: $I = \int \frac{(t/2) e^t}{(1+t)^2} \cdot \frac{dt}{2} = \frac{1}{4} \int \frac{t e^t}{(1+t)^2} dt$. We can rewrite the numerator $t$ as $(t+1-1)$: $I = \frac{1}{4} \int e^t \left( \frac{t+1-1}{(1+t)^2} \right) dt = \frac{1}{4} \int e^t \left( \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$. Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = \frac{1}{1+t}$ and $f'(t) = -\frac{1}{(1+t)^2}$: $I = \frac{1}{4} e^t \left( \frac{1}{1+t} \right) + C$. Substituting $t = 2x$ back: $I = \frac{e^{2x}}{4(1+2x)} + C$.

Evaluate the integral: $\int \frac{x e^{2x}}{(1+2x)^2} dx = $ (where $C$ is a constant of integration.)

Similar Questions

$\int e^x \cdot \sec x(1+\tan x) \, dx = $ . . . . . . $+ C$.

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

$\int e^{4x}(\sin 3x - \cos 3x) dx = $

Let $f(x) = \int \frac{(2-x^2)e^x}{(\sqrt{1+x})(1-x)^{3/2}} dx$. If $f(0) = 0$,then $f(\frac{1}{2})$ is equal to:

If $\int e^{\sin x} \cdot \left[ \frac{x \cos^3 x - \sin x}{\cos^2 x} \right] dx = e^{\sin x} f(x) + c$,where $c$ is the constant of integration,then $f(x)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module