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How to find the integral of x^3 e^(x^2) / (x^2 + 1)^2 ?

  • Thread starter rocomath
  • Start date
  • #1
1,752
1
Ok wow, I can't solve this :(

[tex]\int\frac{x^3e^{x^2}}{(x^2+1)^2}dx[/tex]

I have let u equal almost everything possible.

[tex]u=x^2[/tex]
[tex]d=2xdx[/tex]

[tex]u=e^{x^2}[/tex]
[tex]\ln u=x^2[/tex]
[tex]\frac{du}{u}=2xdx[/tex]

[tex]u=x^2+1 \leftrightarrow u-1=x^2[/tex]
[tex]\ln u=\ln{(x^2+1)}[/tex]
[tex]\frac{du}{u}=\frac{2xdx}{x^2+1}[/tex]
 

Answers and Replies

  • #2
3
0
rocomath, is the integral given as a definite integral? It is possible to find the value of the definite integral by using integration by parts (think about the change of variables a bit). I don't really see an easy u-substitution that's going to help you out, but maybe I'm just blind.
 
  • #3
1,752
1
OH, lol. I forgot to include what dV was ...

But yes, I've been trying parts on it for a while now ... still nothing!
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,832
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Hi rocomath! :smile:

Go with [tex]u=x^2[/tex] again …

you should get [tex]\int\frac{ue^u}{(1 + u)^2}du[/tex]

Then use partial fractions, and try the obvious. :smile:
 
  • #5
1,752
1
Hi rocomath! :smile:

Go with [tex]u=x^2[/tex] again …

you should get [tex]\int\frac{ue^u}{(1 + u)^2}du[/tex]

Then use partial fractions, and try the obvious. :smile:
Uh!!! I didn't even think of using Partial Fractions bc the people in the group said they weren't doing that yet ... so frustrated, lol.

tiny-tim you're a life saver!
 
  • #6
1,752
1
Ok ... so I have

[tex]\frac 1 2\int\frac{te^t}{(t+1)^2}dt[/tex]

Now using Partial Fractions and leaving the constant 1/2 out for now:

[tex]\frac{te^t}{(t+1)^2}=\frac{A}{t+1}+\frac{B}{(t+1)^2}[/tex]

[tex]te^t=A(t+1)+B[/tex]

if t = -1

[tex]B=-\frac 1 e[/tex]

I'm having a hard time solving for A ...

I know what B is, so if t = 0

[tex]A=-\frac 1 e[/tex] ???

... hmm, definitely not right!

I've never done a problem that had a product on the left side.
 
Last edited:
  • #7
3
0
The method of partial fractions won't work like it normally does here since you can't find integers [itex]A[/itex] and [itex]B[/itex] such that

[tex]t e^t = A (t + 1) + B[/tex]

You can, however, solve for [itex]A[/itex] and [itex]B[/itex] in terms of your variable [itex]t[/itex], but it doesn't really make the problem any easier. We might find

[tex]B = - e^{-1}[/tex]

And so plugging that in to the equation above we find

[tex]A = \frac{t e^t - e^{-1}}{t + 1}[/tex]

I still suggest trying the method of integration by parts. Hint: can you get the entire denominator [itex](x^2+1)^2[/itex] into [itex]dv[/itex] such that you can easily integrate it?
 
  • #8
1,752
1
Ok, got it!

Uh, sucha killer. Thanks for the hint Acetv, whewww!!!
 
  • #9
Gib Z
Homework Helper
3,346
4
We still could have used ordinary partial fractions - I'm not sure why you thought it was even possible to find constants A and B for te^t = A(t+1) + B, but tiny-tims suggestion was to use partial fractions on what you've been taught to use it on - Rational functions.

IE Only the [tex]\frac{t}{(t+1)^2}[/tex]. After that, multiply each term by e^t.

You should have got [tex] \frac{e^t}{t+1} - \frac{e^{t}}{(t+1)^2} = \frac{1}{e} \left( \frac{e^u}{u} - \frac{e^u}{u^2} \right) [/tex] where u = t+1, which are simple to integrate.
 
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