- #1

inquisitive

- 2

- 0

[itex]

\int du/(u^2-a^2) = (1/2a) ln (u+a)/(u-a) + C

[/itex]

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- Thread starter inquisitive
- Start date

- #1

inquisitive

- 2

- 0

[itex]

\int du/(u^2-a^2) = (1/2a) ln (u+a)/(u-a) + C

[/itex]

- #2

mathwonk

Science Advisor

Homework Helper

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haveyou tried taking the derivative of the RHS?

- #3

swapnilster

- 10

- 0

=>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

Big-T

- 64

- 0

You could also do this by partial fraction decomposition.

- #5

DavidWhitbeck

- 351

- 1

[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

inquisitive

- 2

- 0

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.

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

- 43,021

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