# evaluate an integral

• May 13th 2013, 07:47 AM
dumbledore
evaluate an integral
How do I compute this integral?

$\displaystyle \int \frac{e^\sqrt5x}{3x} dx$
• May 13th 2013, 08:03 AM
Prove It
Re: evaluate an integral
Quote:

Originally Posted by dumbledore
How do I compute this integral?

$\displaystyle \int \frac{e^\sqrt5x}{3x} dx$

Is it \displaystyle \displaystyle \begin{align*} \int{\frac{e^{\sqrt{5}\,x}}{3x}\,dx} \end{align*} or \displaystyle \displaystyle \begin{align*} \int{\frac{e^{\sqrt{5x}}}{3x}\,dx} \end{align*}?
• May 13th 2013, 08:09 AM
dumbledore
Re: evaluate an integral
Second one - sorry about that
• May 13th 2013, 08:22 AM
Prove It
Re: evaluate an integral
\displaystyle \displaystyle \begin{align*} \int{\frac{e^{\sqrt{5x}}}{3x}\,dx} &= \frac{1}{3} \int{ \frac{e^{\sqrt{5}\,\sqrt{x}}}{\sqrt{x}\,\sqrt{x}} \, dx } \\ &= \frac{ 2 }{ 3 \, \sqrt{5} } \int{ \frac{ \sqrt{5} \, e^{ \sqrt{5} \, \sqrt{x} } }{ 2 \, \sqrt{x} \, \sqrt{x} } \,dx } \end{align*}

Now make the substitution \displaystyle \displaystyle \begin{align*} u = \sqrt{5} \, \sqrt{x} \implies du = \frac{ \sqrt{5} }{ 2 \, \sqrt{x} } \, dx \end{align*} and the integral becomes

\displaystyle \displaystyle \begin{align*} \frac{ 2 }{ 3 \, \sqrt{5} } \int{ \frac{ \sqrt{5} \, e^{ \sqrt{5} \, \sqrt{x} } }{ 2 \, \sqrt{x} \, \sqrt{x} } \,dx } &= \frac{ 2 }{ 3 \, \sqrt{5} } \int{ \frac{ e^u }{ \frac{ u }{ \sqrt{5} } } \, du } \\ &= \frac{2}{3} \int{ \frac{e^u}{u} \,du} \end{align*}

and this can only be evaluated in terms of the Exponential Integral. Have you encountered this before?
• May 13th 2013, 09:16 AM
dumbledore
Re: evaluate an integral
No I have not encountered this before. Is there any other way of solving the integral?