Integral Proof: u^2-a^2 | Detailed Explanation

[inquisitive]

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

 

[mathwonk]

haveyou tried taking the derivative of the RHS?

 

[swapnilster]

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

 

[Big-T]

You could also do this by partial fraction decomposition.

 

[DavidWhitbeck]

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

You need to show that you've attempted the problem.

 

[inquisitive]

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
<br /> <br /> \frac{1}{2}ln \frac{(u+a)}{\sqrt{u^2-a^2}}+ C<br /> <br />
This is where I started to need some help from you guys. But after some time I realized to arrive to the form
<br /> <br /> \frac{1}{2a}ln \frac{u+a}{u-a}+ C;<br /> <br />
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.

 

[HallsofIvy]

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".