I've spent about a good 30 minutes on this problem and I can't seem to find a way to use u-sub to make this problem work

(x + 1)e^(-3lnx)

Any help is appreciated! (Crying)

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- Jun 14th 2008, 10:12 AMJonathanEyoonIntegrating with U-Substitution
I've spent about a good 30 minutes on this problem and I can't seem to find a way to use u-sub to make this problem work

(x + 1)e^(-3lnx)

Any help is appreciated! (Crying) - Jun 14th 2008, 10:20 AMKrizalid
hence

- Jun 14th 2008, 10:22 AMJonathanEyoon
HUH?

How did it go from there to 1 / x^3? (Surprised) - Jun 14th 2008, 10:24 AMMathstud28
- Jun 14th 2008, 10:32 AMJonathanEyoon
Ok i think I see what happened here but I have a question.

When it got to

e^ln(x^-3)

What you did was bring the ln(x^-3) down by taking the ln to do so right? But wouldn't that end up being

ln(x^-3) lne since you have to do ln of something to bring down whatever is in the exponent? Sorry for the stupid questions =/ just really want to understand it. - Jun 14th 2008, 10:39 AMMathstud28
- Jun 14th 2008, 10:40 AMSilver
- Jun 14th 2008, 10:44 AMJonathanEyoon
OoHhhH I totally forgot that e and ln were inverse functions!! *sigh* I must remember this!!

So for instance another question that is similar to the one posted above is

(x^2 + 2x + 1)e^(-ln(x - 1)) dx

is equal to

(x^2 + 2x + 1) / (x - 1) dx

right? - Jun 14th 2008, 10:46 AMMathstud28
- Jun 14th 2008, 10:47 AMSilver
- Jun 14th 2008, 10:51 AMJonathanEyoon
I appreciate the help guys~!