Evaluate $\int xe^{\frac{3ex^2}{2}}\operatorname d\!x$

Evaluate $$\int xe^{\frac{3 e x^2}{2}}\operatorname d\!x$$

The answer is $\frac{xe^{\frac{3ex^2}{2}}}{3e}+c^{te}$.

But how do I get here? Does anyone have any pointers or a good website which explains this in detail? Thanks.

Thanks for all the input but I'm still not sure how to solve this. My approach is the following:

Let's say I start with $\int xe^t \! dx$. How do I get to $\frac{e^t}{3e}$? When using the integration by parts explanation at mathisfun I get the following:

• $\int xe^t \; dx$
• $u: x \rightarrow u'=1$
• $v: e^t \rightarrow \int v = e^t$
• $\int u v \; dx = u \; \int v \; dx - \int u'(\int v \; dx) \;dx$
• So: $\int xe^t \; dx = x * e^t - \int 1 * e^t \; dx$
• Is: $x * e^t - e^t + C$

Which is not the answer I'm looking for.

• I suppose that you noticed that $x$ looks like the derivative of $x^2$ which appears in the exponential. – Claude Leibovici Feb 23 '14 at 10:00
• @Bart what is t in the term $c^{te}$ – happymath Feb 23 '14 at 10:00
• $constante$ I bet – Claude Leibovici Feb 23 '14 at 10:02
• Guess and check. It is natural to start by differentiating the whole thing or the whole thing except the x factor. You see one of these is off only by a constant multiplier and you fix the constant. – Jacob Wakem Feb 24 '14 at 21:15

I am afraid that, if the integral you wrote is correct, the answer you report is wrong.

To make the problem simpler, change variable $\frac{3 e x^2}{2}=t^2$; so $x=\sqrt{\frac{2}{3 e}} t$ and $dx=\sqrt{\frac{2}{3 e}} \text{dt}$. Replacing all of that leads to the integrand $$\frac{2}{3e} t e^{t^2}$$ and, please notice that $2t$ is the derivative of $t^2$. So, the integral is $$\frac{e^{t^2}}{3 e}$$ Replace now $t^2$ by $\frac{3 e x^2}{2}$ and you end with the result $$\frac{e^{\frac{3 e x^2}{2}}}{3 e} + C$$ which is the correct answer.

You could check easily that the derivative of the "answer" you posted does not match the integrand. So, there is a typo somewhere.

What suggested user129901 (I did not see his answer and I really apologize) is even simpler since, using his change of variable, your integrand becomes simpler as $$\frac{e^t}{3 e}$$

• Thanks for the explanation, but I still don't understand how to get to the answer. I've updated my answer to reflect the steps in my reasoning. Additional help is greatly appreciated :) – Bart Feb 24 '14 at 21:06

Hint:

Use the following substitution $\dfrac{3ex^2}{2}=t$ and proceed.

• I sincerely apologize ! I did not see your answer. Your solution is simpler. I updated my answer to acknowledge. Cheers. – Claude Leibovici Feb 23 '14 at 10:43
• @ClaudeLeibovici there is no need to apologize having multiple answers for the same question is better -). – happymath Feb 23 '14 at 10:47

Here's how I approached it:

Hmmm... Integral of e raised to a funny power of x, then multiplied by a funny function of x, followed by dx. I haven't a clue!. The only thing I can integrate is $e^x dx$. I am so scre...

Wait; I can integrate $e^u du$, where $u$ is any weird and wonderful function of x, and $du$ is the differential of that weird and wonderful function of x. Could it be? Am I saved?

Well, $u$ would have to be defined by:$$u=\frac{3ex^2}{2}$$and then $$du=\frac{du}{dx}dx=3ex\,dx$$Now, 99+% of the time, this would not lead anywhere, but here, thank Goodness, it works (as it often does in Calc 101 textbook questions). I can put a $\frac{1}{3e}$ outside the integral, and $3e$ inside the integral, and I've turned the ugly integral into a constant times the only exponential integral I can do!!

Use the substitution $\frac{3e}{2} x^2 = t^2$ and note that $3ex \mathrm{d}x = 2t\mathrm{d}t$.

Then you will see that $x \mathrm{d}x = \frac{2t}{3e} \mathrm{d}t$

$$\int \! xe^\frac{3ex^2}{2} \, \mathrm{d}x = \frac{1}{3e} \int \! 2te^{t^2} \, \mathrm{d}t$$

Since $2t =$ the derivative of $t^2$:

$$\frac{1}{3e} e^{t^2} + C = \frac{1}{3e}e^{\frac{3ex^2}{2}} + C$$