Hello,
I would like to solve the following integral
For and for
I tried some substitutions without success.
Could anybody help me solving this?
Regards,
Mr. Fogg
Hi
Your problem is the same as solving this integral
In your problem the discriminant is negative if and positive if
This integral can be solved by parts and then by rearranging a little bit into
The last integral depends by the sign of the discriminant
If negative then you can chose the proper change of variables to get (1+u²) at the denominator, and then integrate into arctan
If positive then you can factor the denominator into 2a²(x-x0)(x-x1) and go further in the decomposition of the last integral
So far I did not check the second case
But for the first case I applied the formula I gave you in my previous posts and I could find the good result
In your problem the discriminant is negative if and positive if
And
Since you are integrating between 0 and 1 you do not have to care about the first term because for 1, ln(ax²+bx+c) gives ln(a+b+c)=ln(1)=0 ; and for 0, ln(ax²+bx+c) gives ln(c)=ln(1)=0
The second term (-2x) gives -2 for 1 and 0 for 0, therefore -2
For the third term
For 1,
For 0,
But Arctan(-x)=-Arctan(x) therefore for 0 it gives
Therefore the third term gives
At the end the third term gives