Find the integral int_0^1 1/(5+2x-2x^2) (1+e^(2-4x))dx

  • Thread starter FedEx
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  • #1
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[tex]\int_0^1\frac{1}{(5+2x-2x^2)(1+e^{2-4x})}dt
[/tex]

Believe me. I tried this sum for a good 1 hour.

Tried Parts. All types of substitution. Almost every known method taught to a student of 12th grader. But in vain. A storke of genious. Thats what i need to solve this.

I took the entire denominator for substituion. Just the polynomial. Just the exponential term. In parts i took just the polynomial. Just the exponential term. Everything. No benefits.
 

Answers and Replies

  • #2
CompuChip
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And all that while the answer is just
[tex]\frac{1}{(5+2x-2x^2)(1+e^{2-4x})}[/tex].
 
  • #3
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[tex]dt[/tex] not [tex]dx[/tex]. :shy: Foolish to do that to students.
 
  • #4
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Never expected that from you. Its abvious that its a typing mistake. Its dx.

PS I solved the integral by dirk_mec1. Have a look.
 
  • #5
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Hey compuchip...

Sorry. I think i was quite arrogant in the last post.

My apologies.
 
  • #6
CompuChip
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Hi FedEx,

No need to apologize, my reply wasn't very helpful either :smile:
And it's very good that you solved the dirk_mec's integral, although I hope you'll excuse me for not working through the whole calculation :tongue: That I happen to have this label besides my name by no means implies that I know all the answers or I'm good at everything.

As for your question: I did the integration numerically, and it doesn't look particularly nice. In fact I would really see any good way of solving it, which makes sense since you indicated you have tried all common methods already. I wonder if there is a closed form solution...
 
  • #7
318
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Hard luck. Theres no solution. I saw this sum on an IIT aspirants website. Unsolved. But i am damn sure that even the toughest of IIT Papers doesnot have a sum like this.
 

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