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

Oneiromancy

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Sep 28, 2007
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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.
 
Oneiromancy said:
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:

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

du=3213x+9dx\displaystyle du \, = \, \frac{3}{2}\frac{1}{\sqrt{3x+9}} \, dx

dx=23udu\displaystyle dx \, = \, \frac{2}{3}\cdot u \, du

That should result in a more familiar integrand.
 
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