Integration by Parts

Hi, guys. I'm having trouble with this one problem. You have to use integration by parts, and this one is coming out very messy for me. I doubt I'm even doing this correctly.

integral of: ln((x^2)-1)dx

so, u = ln((x^2)-1)
u' = 2x/((x^2)-1)
v = x
v' = 1

I'm guessing those are my best choices.

I better not be making this more complicated than necessary.

This is calculus, right?

This is calculus, right?

Yep. Calculus 2.

I start that either this quarter or next quarter. Aw, yeah! If this was just an integral I could do it.

Oh man I haven't done this stuff in like 3 years. I'll need this to refresh for physics eventually I suppose...

I just talked to a friend. This requires integration by parts like 3 times... so forget this, everyone. Thanks, anyway!

I'll solve it in the future!

shame max meso isn't around.

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)...

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)...

yeah you did it right

And btw if you need help with another one, use wolfram alpha here just put integrate and the integral with dx at the end.

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...

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...

lol, yeah i didnt saw what kalovale did, i thought that he did algebraic integration at x^2/(x-1)(x+1) dx after setting the by parts method

Mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm fucking calculus nerds getting here at night before I could come look smart...

Don't you people have jobs or school you need to fucking sleep for?

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.

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.

ln(ab) = ln(a) + ln(b)

Just test with a case you know the answer for both sides.

ln(e^2) = ln(e) + ln(e)

Aw. I would have enjoyed solving this one.

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.

oh my bad, line 5 is fine, didn't know where that separation came from.

arctan(x) = int( 1/(x^2 + 1) )
...from memory
...took calc like 5 years ago
...close to math masters

dont need trig for partial fractions

Thanks for all the tips, bros. If I run into more trouble, I'll post 'em here!

Bump. (oops, sorry for the double post)

I am sure I did this problem correctly, but this damn online program isn't accepting it.

I'll just check with you guys to see if I'm an idiot or the program is being stupid.

This is NOT integration by parts, but instead integration by the use of a given table. I hope you guys know what I'm talkin' about!

integral of: ((x^2)-7)/((x^2)-4)dx

using long division: (-3/((x^2)-4))+1

edit: FORGET EVERYTHING, I GOT IT!