Integrate the function: $x^{2} e^{x}$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
Let $I = \int x^{2} e^{x} dx$.
Using the integration by parts formula,$\int u v dx = u \int v dx - \int (u' \int v dx) dx$.
Taking $u = x^{2}$ and $v = e^{x}$,we get:
$I = x^{2} \int e^{x} dx - \int (\frac{d}{dx} x^{2} \cdot \int e^{x} dx) dx$
$I = x^{2} e^{x} - \int 2x e^{x} dx$
$I = x^{2} e^{x} - 2 \int x e^{x} dx$.
Applying integration by parts again for $\int x e^{x} dx$ with $u = x$ and $v = e^{x}$:
$\int x e^{x} dx = x e^{x} - \int (1 \cdot e^{x}) dx = x e^{x} - e^{x}$.
Substituting this back into the expression for $I$:
$I = x^{2} e^{x} - 2(x e^{x} - e^{x}) + C$
$I = x^{2} e^{x} - 2x e^{x} + 2e^{x} + C$
$I = e^{x}(x^{2} - 2x + 2) + C$,where $C$ is an arbitrary constant.

Explore More

Similar Questions

Evaluate the integral: $\int x^3 \log x \, dx$

If $n \in N$ and $I_{n}=\int(\log x)^{n} dx$,then $I_{n}+n I_{n-1}$ is equal to

If $\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=f(x)-\log \left(1+x^2\right)$, then $f(x)$ is equal to

$\int \cos 2\theta \log \left( \frac{\cos \theta + \sin \theta }{\cos \theta - \sin \theta } \right) d\theta = $

$I = \int \cos(\ln x) \, dx$. Then $I =$

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