Indefinite integral

  • Thread starter dapias09
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  • #1
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Main Question or Discussion Point

Hello, I need help with a very hard integral, I was trying several steps and I tried in the software "derive" but the answer didn't like to me.

It is ψ(x,t)= [b^2-(x-vt)^2]^(-2) . I must integrate it with respect to x .

Thanks in advance for any help!

PD: I tried x-vt = u like the first step du=dx. My teacher adviced to me continue with [b^2-u^2]^(-2) = z but I don't know how to continue in the best way.
 

Answers and Replies

  • #2
798
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My teacher adviced to me continue with [b^2-u^2]^(-2) = z but I don't know how to continue in the best way.
Find A, B, C , D so that :
1/(b²-u²)² = A/(b+u) +B/(b-u) +C/(b+u)² +D/(b-u)²
Then you can easily integrate.
 

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