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Calculus Help

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guissmo
Hey, I need help with this problem. I'm about to have an exam in less than 24 hours and our professor gave us this problem. I'm not exactly sure if it's "solvable" but the question he gave is:

What is the integral of xe^xlnxdx?

If you have other resources that can help me study for Integration by Parts, Reduction Formula, Trig Substitutions, Partial Fractions and z = tan(x/2), please post it here. Thanks. Smile
infinisa
Hi guissmo

You want to find ∫x e^x lnx dx

Let's start with an easier problem:

∫x e^x dx
We solve by integration by parts, using
∫u v' dx = uv - ∫u' v dx
and setting u = x (so u' = 1) and v'=e^x (so v = e^x)

we get:
∫x e^x dx = x e^x - ∫1 e^x dx
= x e^x - e^x
= e^x (x - 1)

Now returning to the original problem:
∫x e^x lnx dx

We solve by integration by parts Using again
∫u v' dx = uv - ∫u' v dx
and setting u = ln x (so u' = 1/x) and v'=x e^x (so v = e^x (x - 1) by the work above)

we get:
∫(ln x) x e^x dx = (ln x) (e^x (x - 1)) - ∫1/x (e^x (x - 1)) dx
= (ln x) (e^x (x - 1)) - ∫e^x (1 - 1/x)) dx
= (ln x) (e^x (x - 1)) - ∫e^x dx + ∫( 1/x) e^x dx
= (ln x) (e^x (x - 1)) - e^x + ∫( 1/x) e^x dx
= e^x ( ( (x - 1)(ln x) - 1) + ∫( 1/x) e^x dx

I'm afraid I got stuck on ∫( 1/x) e^x dx though!

For help on the topics you mention, you might like to look at http://www.ibmaths.com/ .

What exams are you studying for?

Good luck!
Afaceinthematrix
infinisa wrote:
Hi guissmo

You want to find ∫x e^x lnx dx

Let's start with an easier problem:

∫x e^x dx
We solve by integration by parts, using
∫u v' dx = uv - ∫u' v dx
and setting u = x (so u' = 1) and v'=e^x (so v = e^x)

we get:
∫x e^x dx = x e^x - ∫1 e^x dx
= x e^x - e^x
= e^x (x - 1)

Now returning to the original problem:
∫x e^x lnx dx

We solve by integration by parts Using again
∫u v' dx = uv - ∫u' v dx
and setting u = ln x (so u' = 1/x) and v'=x e^x (so v = e^x (x - 1) by the work above)

we get:
∫(ln x) x e^x dx = (ln x) (e^x (x - 1)) - ∫1/x (e^x (x - 1)) dx
= (ln x) (e^x (x - 1)) - ∫e^x (1 - 1/x)) dx
= (ln x) (e^x (x - 1)) - ∫e^x dx + ∫( 1/x) e^x dx
= (ln x) (e^x (x - 1)) - e^x + ∫( 1/x) e^x dx
= e^x ( ( (x - 1)(ln x) - 1) + ∫( 1/x) e^x dx

I'm afraid I got stuck on ∫( 1/x) e^x dx though!

For help on the topics you mention, you might like to look at http://www.ibmaths.com/ .

What exams are you studying for?

Good luck!


I got the exact same answer when I tried this problem last night after he posted it... I got stuck on the integral of (e^x)/x. I do not think that it has any elementary answer. You could try to integrate it with a power series:

e^x = ∑ xⁿ/n!
e^x / x = 1/x + ∑ xⁿ / (n+1)!.

Therefore,
∫e^x / x dx = ln|x| + ∑ xⁿ / (n·n!) + C


But I do not think that that helps out at all in this problem. I tried all the combinations in this problem (first integrating xe^x, then xlnx, (lnx)e^x) and then gave up... there may be some trick but I did not see it... I am sad to say that I failed at this problem... I then went to bed thinking I'd try it again in the morning and then I saw you posted and failed at the same part...


guissmo: I'll post some basic help for the topics you requested...

Integration by Parts: this is when you integrate the product of two functions... You're basically using the product rule.

(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)
Integrate both sides
f(x)g(x) = ∫f'(x)g(x) + ∫f(x)g'(x)
∫f'(x)g(x) = f(x)g(x) - ∫f(x)g'(x)

Then, of course, people replace g with u and f with v, yielding:
∫udv = uv - ∫vdu

If you're like me and you hate memorizing formulas, just remember that integration by parts is obtained by the product rule and then you'll never have to memorize it (of course you'll have to derive it every time you use it...)

For reduction formula, trig substitution, etc. I would just suggest finding examples online or in your book and then if you cannot solve one of the examples posting here... I don't really feel like digging out my old calculus book to search for examples to post here, that's your job.

Partial fractions are easy; it's just making fractions easier... I'll post an example off the top of my head.

∫1/(x^2+4x+3)dx
1/(x^2+4x+3) = 1/((x+3)(x+1)) = A/(x+3) + B/(x+1)
Multiply both sides by x+3
1/(x+1) = A + B(x+3)/(x+1)
Choose a convienent value of x (x=-3 so that B goes away)
-1/2 = A
Do the same for B (multiply both sides by x+1
1/(x+3) = A(x+1)/(x+3) + B
Let x = -1
1/2 = A

Therefore ∫1/(x^2+4x+3)dx = (-1/2)∫1/(x+3)dx + (1/2)∫1/(x+1)dx
∫1/(x^2+4x+3)dx = (-1/2)ln|x+3| + (1/2)ln|x+1| + C
guissmo
Wow, thanks. I didn't know Frihost would still rock when it comes to studying. Haha.
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