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

[FedEx]

$$\int_0^1\frac{1}{(5+2x-2x^2)(1+e^{2-4x})}dt$$

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.

 

[CompuChip]

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

 

[Ofey]

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

 

[FedEx]

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.

 

[FedEx]

Hey compuchip...

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

My apologies.

 

[CompuChip]

Hi FedEx,

No need to apologize, my reply wasn't very helpful either
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 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...

 

[FedEx]

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.