Question

Show that:
\[\newcommand \dd {\,\mathrm d}
\boxed{
\int x^2e^{ax}\dd x
= \left(\frac{x^2}a - \frac{2x}{a^2} + \frac2{a^3}\right)e^{ax}
}\]

Answer 1

what about integration by parts

Answer 2

\[
\int x^2e^{ax}\dd x
= \frac{x^2}ae^{ax}-\frac{2}a\int xe^{ax}\dd x
\]

Answer 3

\[=\frac{x^2}ae^{ax}-\frac2a\left[\frac xae^{ax}-\frac1a\int e^{ax}\dd x\right]\]

Answer 4

\[=\frac{x^2}ae^{ax}-\frac{2x}{a^2}e^{ax}+\frac2{a^2}\int e^{ax}\dd x\\
=\frac{x^2}ae^{ax}-\frac{2x}{a^2}e^{ax}+\frac2{a^3}e^{ax}+C\\
=\left(\frac{x^2}{a}-\frac{2x}{a^2}+\frac2{a^3}\right)e^{ax}+C\]

Answer 5

you can also differentiate both sides to prove that one thing is equal to the other

Answer 6

we may try differentiation under the integral :

\[\int x^2e^{ax}\,dx=\int \dfrac{\partial^2}{\partial a^2}e^{ax}\,dx\]

Answer 7

\[\text{we can say } \int\limits f(x) dx=g(x) \text{ by verifying } f(x)=g'(x)\]

Answer 8

More generally :

\[\int x^ne^{ax}\,dx=\int \dfrac{\partial^n}{\partial a^n}e^{ax}\,dx = \dfrac{\partial^n}{\partial a^n} \int e^{ax}\, dx = \dfrac{\partial^n}{\partial a^n} \frac{e^{ax}}{a}\]

Answer 9

\[\frac{\dd}{\dd x}\left(\frac{x^2}a - \frac{2x}{a^2} + \frac2{a^3}\right)e^{ax}\\
\qquad=\left[\left(\frac{2x}a - \frac{2}{a^2}\right)+\left({x^2} - \frac{2x}{a} + \frac2{a^2}\right)\right]e^{ax}\\
\qquad =x^2e^{ax}\\[4ex]
\qquad \Downarrow\\[2ex]
\int x^2e^{ax}\dd x = \left(\frac{x^2}a - \frac{2x}{a^2} + \frac2{a^3}\right)e^{ax}
\]

Answer 10

\[\begin{align}
\int x^2e^{ax}\dd x
&=\int \frac{\partial^2}{\partial a^2}e^{ax}\dd x\\
&= \dfrac{\dd ^2}{\dd a^2} \int e^{ax}\dd x\\
&= \dfrac{\dd^2}{\dd a^2}\left(\frac{e^{ax}}a\right)\\
&=\frac{\dd }{\dd a}\left(\frac{ax-1}{a^2}e^{ax}\right)\\
&=\frac{a^2\big(x+x(ax-1)\big)-2a\big(ax-1\big)}{a^4}e^{ax}\\
&=\frac{\Big(a^2x+a^3x^2-a^2x\Big)-\Big(2a^2x-2a\Big)}{a^4}e^{ax}\\
&=\frac{a^3x^2-2a^2x+2a}{a^4}e^{ax}\\
&=\left(\frac{x^2}a-\frac{2x}{a^2}+\frac{2}{a^3}\right)e^{ax}\\
\end{align}\]

Answer 11

Thank you @myininaya
Thank you @ganeshie8