फलन का समाकलन कीजिए: $\frac{x e^{x}}{(1+x)^{2}}$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
माना $I = \int \frac{x e^{x}}{(1+x)^{2}} dx = \int e^{x} \left\{ \frac{x}{(1+x)^{2}} \right\} dx$
$= \int e^{x} \left\{ \frac{1+x-1}{(1+x)^{2}} \right\} dx$
$= \int e^{x} \left\{ \frac{1}{1+x} - \frac{1}{(1+x)^{2}} \right\} dx$
माना $f(x) = \frac{1}{1+x}$,तो $f'(x) = -\frac{1}{(1+x)^{2}}$
$\Rightarrow \int \frac{x e^{x}}{(1+x)^{2}} dx = \int e^{x} \{f(x) + f'(x)\} dx$
हम जानते हैं कि $\int e^{x} \{f(x) + f'(x)\} dx = e^{x} f(x) + C$
$\therefore \int \frac{x e^{x}}{(1+x)^{2}} dx = \frac{e^{x}}{1+x} + C$
जहाँ $C$ एक स्वेच्छ अचर है।

Explore More

Similar Questions

$\int {{e^{2x}}\frac{{1 + \sin 2x}}{{1 + \cos 2x}}} \,dx = $

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

निश्चित समाकलन $\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x$ का मान ज्ञात कीजिए।

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

$\int \left( \frac{1-\log x}{1+(\log x)^2} \right)^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