indefinite integral with exponential

How I can calculate the following indefinite integral $$\int \frac{e^{\sqrt{2x-1}}}{e^{3x}}dx?$$

I try integration by parts but it not seems be useful.

Can someone give me a hint? Thanks for advance.

1 Answer

\begin{align} \int \frac{e^{\sqrt{2x-1}}}{e^{3x}}dx &= \int e^{\sqrt{2x-1}-3x}dx\\ u=2x-1, du = 2dx \quad&=2\int e^{\sqrt{u}-\frac{3}{2}(u+1)}du\\ t^2 = u, du = 2tdt \quad &=4 \int t e^{t-3(t^2+1)/2}dt\\ &=4\int t e^{-3t^2/2+t-3/2}dt\\ &=4 \int t e^{-3/2\left((t-1/3)^2+8/9\right)}dt\\ &=4e^{-4/3}\int te^{-3/2(t-1/3)^2}dt\\ t-1/3 = v \quad&= 4e^{-4/3} \int (v+1/3)e^{(-3/2)v^2}dv\\ &=4e^{-4/3} \left( \int ve^{(-3/2)v^2}dv + \int (1/3)e^{(-3/2)v^2}dv\right) \end{align} I am not sure how to procede with the indefinite integral of the second term, but it is quite close to the Error function definite integral. The first is a simple u-substitution. I'm not sure if I'm right so far because I can't check the mathematica answer at the moment unfortunately.

• It's (up to multiplication by a constant) the pdf of the normal distribution, and the indefinite integral has no closed form. Jan 22 '18 at 19:48
• @GNUSupporter what is a pdf? Jan 22 '18 at 23:22
• probability density function Jan 23 '18 at 0:01
• @GNU Supporter I have zero clue what that is, so if you'd like to finish the answer feel free to copy my process :-) Jan 23 '18 at 0:17
• You can continue integrating the second term as $\operatorname{erf}(x)$ Jan 23 '18 at 0:23