Integration: e^sqrt(3x+9)dx

[Oneiromancy]

My book said to do a substitution prior to integrations by parts. I tried this:

u = 3x + 9
du = 3dx

= (1/3)integral e ^ [(u)^(1/2)]

v = u^(1/2)
dv = 1/[2u^(1/2)]
dw = du
w = u

I dunno, that didn't get me anywhere because I didn't know what to do with the e.

 

[Deleted member 4993]

My book said to do a substitution prior to integrations by parts. I tried this:

u = 3x + 9
du = 3dx

= (1/3)integral e ^ [(u)^(1/2)]

v = u^(1/2)
dv = 1/[2u^(1/2)]
dw = du
w = u

I dunno, that didn't get me anywhere because I didn't know what to do with the e.

Make the following substitution in stead:

\(\displaystyle u \, = \, \sqrt{3x+9}\)

\(\displaystyle du \, = \, \frac{3}{2}\frac{1}{\sqrt{3x+9}} \, dx\)

\(\displaystyle dx \, = \, \frac{2}{3}\cdot u \, du\)

That should result in a more familiar integrand.