Mibs Wrote:This is calculus, right?

Kalovale Wrote:Hope I did it right, lol.

EDIT: I meant "let v' = 1"... or maybe I didn't... idk, the convention seems silly to me. I never do it this way.

I just realized this would've been super easy if you let ln(x^2-1) = ln(x-1) + ln(x+1)...

modular Wrote:you are correct up to the point where you write out uv - int(vdu)

to evaluate int( x^2 / ( x^2 - 1 ) ), use partial fractions to split the bottom into linear terms, do a u substitution to move the 1's into the numerator instead of the denominator, and then evaluate the resulting simple integrals. but DO NOT split or multiply integrals the way you did in lines 5 & 6.

very tedious, very torturous, i wrote down about half the work and don't have a math program / word 2010 on this computer. diligence is all you need.

 Spoiler

final answer should be:
x ln(x^2 -1) - 2x + ln(x-1) - ln(x+1) + c

whoa, the ninjas are out early today...

modular Wrote:you are correct up to the point where you write out uv - int(vdu)

to evaluate int( x^2 / ( x^2 - 1 ) ), use partial fractions to split the bottom into linear terms, do a u substitution to move the 1's into the numerator instead of the denominator, and then evaluate the resulting simple integrals. but DO NOT split or multiply integrals the way you did in lines 5 & 6.

 Spoiler

 answer

first red part: xln(x^2 - 1)
second red part: -2x
blue part: -2*int(1/(x^2-1))dx = -2 * (ln(x-1) - ln(x+1))/2 = -ln(x-1) + ln(x+1)

Kalovale Wrote:Why not? Line 6 is blatantly wrong (and I'm ashamed for it) but line 5? Technically int(a+b)dx = int(a)dx + int(b)dx, or am I forgetting something and making pomegranate up on the fly?

 Spoiler

Also, what the hell @ the last line, I totally turned a multiplication into an addition. Serves me right for doing Math without pencil and paper at midnight.

If anything, the last 3 lines are wrong. But until that point, I'm fairly sure I got it right.

 answer

first red part: xln(x^2 - 1)
second red part: -2x
blue part: -2*int(1/(x^2-1))dx = -2 * (ln(x-1) - ln(x+1))/2 = -ln(x-1) + ln(x+1)

And if I'm not hallucinating again, int (1/x^2-1)dx should be one of the hyperbolic integrands or its inverse. Nobody remembers that crap anyway, so I'd stick with integration by parts.