Prove the integral

  • #1

Main Question or Discussion Point

Anyone please help I need a detailed proof of this integral
[itex]
\int du/(u^2-a^2) = (1/2a) ln (u+a)/(u-a) + C
[/itex]
 

Answers and Replies

  • #2
mathwonk
Science Advisor
Homework Helper
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haveyou tried taking the derivative of the RHS?
 
  • #3
You get the answer by taking u=a*sec y =>y=arcsec u/a(arcsec-sec inverse)
=>du=a sec y tan y dy
Therefore I=integral[(a sec y tan y dy)/a^2 tan^2 y]
=>(sec y dy)/(a tan y)
=>cosec y dy /a
=>ln(cosec y-cot y)/a
then just substitute value of y to get the answer
 
  • #4
64
0
You could also do this by partial fraction decomposition.
 
  • #5
Anyone please help I need a detailed proof of this integral
[itex]
\int du/(u^2-a^2) = (1/2a) ln (u+a)/(u-a) + C
[/itex]
You need to show that you've attempted the problem.
 
  • #6
Thanks guys...
The solution swapnilster proposed is correct. I also did his solution before. But after the trigonometric substitution I was stuck at the answer
[itex]

\frac{1}{2}ln \frac{(u+a)}{\sqrt{u^2-a^2}}+ C

[/itex]
This is where I started to need some help from you guys. But after some time I realized to arrive to the form
[itex]

\frac{1}{2a}ln \frac{u+a}{u-a}+ C;

[/itex]
I just need to put the (u+a) term inside the radical and apply some algebra and properties of logarithms. But anyway, thanks again guys.
 
Last edited:
  • #7
HallsofIvy
Science Advisor
Homework Helper
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As pointed out by MathWonk, a perfectly valid method of proving that a given function is an anti-derivative is to differentiate it. That, typically, is easier than trying to integrate "from scratch".
 

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