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

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The integral int_0^1 1/(5+2x-2x^2)(1+e^(2-4x))dx presents significant challenges for standard methods of integration. Forum participants, including users like dirk_mec1 and FedEx, attempted various techniques such as substitution and integration by parts without success. Ultimately, it was concluded that there is no known closed-form solution for this integral, which has been noted as unsolved on IIT aspirants' platforms. The integral's complexity suggests it may not be suitable for typical examination settings.

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FedEx
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\int_0^1\frac{1}{(5+2x-2x^2)(1+e^{2-4x})}dt<br />

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.
 
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And all that while the answer is just
\frac{1}{(5+2x-2x^2)(1+e^{2-4x})}.
 


dt not dx. :shy: Foolish to do that to students.
 


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.
 


Hey compuchip...

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

My apologies.
 


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 :-p 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...
 


Hard luck. there's 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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