વિધેયનું સંકલન કરો: $\frac{x e^{x}}{(1+x)^{2}}$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
ધારો કે $I = \int \frac{x e^{x}}{(1+x)^{2}} dx = \int e^{x} \left\{ \frac{x}{(1+x)^{2}} \right\} dx$
$= \int e^{x} \left\{ \frac{1+x-1}{(1+x)^{2}} \right\} dx$
$= \int e^{x} \left\{ \frac{1}{1+x} - \frac{1}{(1+x)^{2}} \right\} dx$
ધારો કે $f(x) = \frac{1}{1+x}$,તો $f'(x) = -\frac{1}{(1+x)^{2}}$
$\Rightarrow \int \frac{x e^{x}}{(1+x)^{2}} dx = \int e^{x} \{f(x) + f'(x)\} dx$
આપણે જાણીએ છીએ કે $\int e^{x} \{f(x) + f'(x)\} dx = e^{x} f(x) + C$
$\therefore \int \frac{x e^{x}}{(1+x)^{2}} dx = \frac{e^{x}}{1+x} + C$
જ્યાં $C$ એ સ્વૈર અચળાંક છે.

Explore More

Similar Questions

ધારો કે $f(t) = \int \left( \frac{1 - \sin(\ln t)}{1 - \cos(\ln t)} \right) dt$, $t > 1$ માટે. જો $f(e^{\pi/2}) = -e^{\pi/2}$ અને $f(e^{\pi/4}) = \alpha e^{\pi/4}$ હોય, તો $\alpha$ ની કિંમત શોધો.

$\int e^x \left( \frac{x-1}{x^2} \right) dx =$

$\int \left( {1 + x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}}\,dx = $

$\int_1^{e} \frac{e^x}{x}(1+x \log x) d x=$

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

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo