Calculus Integration: partial fractions versus sum of 2 fractions

rayfodder

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Sep 29, 2013
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Hi, I'm trying to figure out what's wrong here if anything.

∫(x+2)/(x^2 - 1)

by solving with partial fractions I get integrals: 3/(2(x-1)) - 1/(2(x+1))

and then the result: (3/2)(ln(x-1)) - (1/2)(ln(x+1)) + C

but solving by first rewriting the problem as: x/(x^2 - 1) + 2/(x^2 - 1)

gives the result: (1/2)ln|x^2 - 1| + ln|x-1| - ln|x+1| + C

Please help explain if something is wrong or if the two answers look different but are the same.
 
Hi, I'm trying to figure out what's wrong here if anything.

∫(x+2)/(x^2 - 1)

by solving with partial fractions I get integrals: 3/(2(x-1)) - 1/(2(x+1))

and then the result: (3/2)(ln(x-1)) - (1/2)(ln(x+1)) + C

but solving by first rewriting the problem as: x/(x^2 - 1) + 2/(x^2 - 1)

gives the result: (1/2)ln|x^2 - 1| + ln|x-1| - ln|x+1| + C

Please help explain if something is wrong or if the two answers look different but are the same.

(1/2)ln|x^2 - 1| + ln|x-1| - ln|x+1| + C = (1/2)[ln|x+1| + ln|x-1|] + ln|x-1| - ln|x+1| + C = ... and continue...
 
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