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.


\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.


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