Can log_e(x) be integrated wrt x? ******** A Problem Worthy of Attack Proves It''s Worth by Fighting Back

Yes.

i''ve got a feeling the answer is 1/x but i could be completely off....


ho hum......




+ + +

x*ln(x) - x + C

yeah, you do it by integration by parts. set u=ln x, dv=dx, so you get

==uv - (integral) v du

==x ln x - (integral) (x/x)dx <-- since du/dx is 1/x dx and v=x

==x ln x - (integral) 1 dx

==x ln x - x

hence what stdio got...and a good refresher for how i worked it out back in calc 2. hope i did it right anyway

[edited by - draqza on May 20, 2003 10:09:47 AM]

Cheers.

As a follow up question, my maths text mentions integrals of inverse circular and hyperbolic functions, but offers no solutions. Can they be integrated?

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A Problem Worthy of Attack
Proves It''s Worth by Fighting Back

Inverse hyperbolic functions, or plain hyperbolic functions?

Plain hyperbolic are easy, A sinh Bx -> A/B cosh Bx, A cosh Bx -> A/B sinh Bx.

Inverse hyperbolic functions are trickier, and use a standard set of results, which from my formula book is:

cosh^-1 (x / a) dx = 1 / root (x^2 - a^2)
sinh^-1 (x / a) dx = 1 / root (x^2 + a^2)

Whose accuracy, of course, is totally dependent on my interpretation of my formula book, which is slightly hard to follow :-)

HTH.

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This is the tale of a Northern Soul, looking to find his way back home

There are vast tables of integrals published. Search google.

no wise fish would go anywhere without a porpoise - The Mock Turtle

MDI, that''s differentiating. Sorry to nitpick but integration is a little different.

Ive tried googling, no luck. ''m wondering if they actually can be integrated perfectly at all.

Are there infinite series defined for the inverse circular and hyperbolic functions?

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A Problem Worthy of Attack
Proves It''s Worth by Fighting Back

AS for ln x you integrate arcsin x and arccos x by parts since you already know how to differentiate arcsin and arccos.
You get for arcsin x:

x arcsin x + root(1 - x^2) + C

I''ll leave arccos x as an exercise for you!