Help integrating a function

[euclidean]

https://www.physicsforums.com/attachments/67272

I need some help, thank you:-)

 

[euclidean]

£(x^(1/2))/(x-1)dx ,the upper limit is 4, the lower limit is 3

 

[SteamKing]

£(x^(1/2))/(x-1)dx ,the upper limit is 4, the lower limit is 3

What exactly does this mean? Why are you trying to integrate Pounds sterling?

 

[euclidean]

What exactly does this mean? Why are you trying to integrate Pounds sterling?

II'm sorry "£" here represent the integration symbol ...I can't type that on my phone...

 

[epenguin]

You've got this awkward square root so a natural way to try and get rid of it would be the substitution y = x² and this seems to bring it back to hopefully familiar or recognisable things.

 

[vanhees71]

I guess the integral is
$$I=\int \mathrm{d} x \frac{\sqrt{x}}{x-1}.$$
Then I'd substitute
$$u=\sqrt{x}, \quad x=u^2 \; \mathrm{d} x = \mathrm{d} u 2 u.$$
Mod note: I removed part of this post as it was too much help.

 

[arildno]

Note that vanhees71's suggestion also suggests a more general thinking on square roots.
Essentially, square roots (not to mention other types of roots!) are nasty, and a good, general procedure is to seek to get rid of them by setting a pesky square root expression equal to a new variable, hoping that your problem disappears, say with productions of squares, rather than square roots.
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In other types of square root problems, the trick is to make a perfect square out of the expression beneath the square root sign, so that this new square precisely cancels out the bothersome root sign.

 

[Mark44]

To the OP: Homework problems need to be posted in the Homework & Coursework sections, not in the technical math sections. I have moved your thread. Please post any future questions in the appropriate section.