Help me w/ a quick calc problem

[20_MuleTeam_Borax]

I need to integrate
e^(3 ln x) dx

I tried doing it using u substitution (whatever it's called . . . where you substitute u for part of the equation, find du, and try to put du in place of dx), but i couldn't figure it out.

 

[Ns1]

n/m i don't wanna even try right now

 

[erikiksaz]

Haha, our class is almost there

 

[Imdmn04]

<< wouldn't that be

e^(3lnx) * (3/x)

i'm on this chapter too so don't quote me haha >>



i think you are taking the derivative

e^(3 ln x) dx=x^3dx

so its 1/4x^4+c when you integrate it

 

[chasem]

well, im in highschool pre-cal

But isnt a natural log, when its e^e, just the natural log in isself,,,

IE:


e^(ln(x))


==

ln(x)


or maybe its just

log(x)


or just


x


i forgot what it equals, i know it equals something though

 

[Ns1]

rofl omfg he said integral

my bad!

 

[yomega]

Simplify, simplify, simplify.

e^(3*ln(x)) dx = e^ln(x^3) dx = x^3 dx

so you get (x^4)/4 + C

EDIT: wow, 3 people posted while I was typing my message , and I forgot that damn constant of integration

 

[xirtam]

I'm with yomega. I think you can simplify e^(3 ln x) to e ^ ((ln x) ^ 3) or x^3. So, integrating x^3 yields x^4/4.

 

[goodoptics]

yep, it's x^4/4 + C

 

[20_MuleTeam_Borax]

bah, god damn algebra incompetence is holding me back