Question

Evaluate the Intergal

Answer 1

\[\int\limits_{}^{}\frac{ 1 }{ x^2-1 }dx \]

Answer 2

I got :

\[\frac{ 1 }{ 2 }(-\ln(x+1)+\ln(x-1)) + C\]

Answer 3

But wolfram gives me an answer of:

\[\frac{ 1 }{ 2 }(\ln(1-x)-\ln(x+1))\]

Answer 4

I don't see why I am wrong...

Answer 5

Why do you think that's wrong? Hast thou forgotten thine algebra? Perhaps you don't recall that \(\int \dfrac{1}{x}\;dx = |x| + C\)?

Answer 6

I don't even havr that anywhere... I used partial fractions.

Answer 7

\[\int\frac{dx}{x^2-1}\]Decompose the integrand into partial fractions:
\[\frac{1}{x^2-1}=\frac{A}{x+1}+\frac{B}{x-1}\\
\text{Solving for A and B, you should get } A = \frac{1}{2} \text{ and }B=-\frac{1}{2}.\]
\[\frac{1}{2}\int\frac{dx}{x+1}-\frac{1}{2}\int\frac{dx}{x-1}, \text{ which gives you}\\
\frac{1}{2}\ln|x+1|-\frac{1}{2}\ln|x-1|+C\\
\frac{1}{2}\left(\ln|x+1|-\ln|x-1|\right)+C\]

Answer 8

Right, and how did you get from the partial fractions to the logarithms?

Answer 9

Allright allright. What is the Integral of :

\[1/(x-1)\]

Answer 10

Isn't it ln|x-1) ?

Answer 11

+C

Answer 12

Notice how the absolute values appear in the excellent demonstratino just provided. Give a certain Domain, the absolute values can be removed.

Answer 13

Ohh I know it's positive because my domain is from 2 to infinity.

Answer 14

Well, there you go. If you copied the Wolfram result exactly, with all assumptions intact, it makes no sense at all. Put x = 3 in there. Bad.

Answer 15

Wait. I found my mistake.

Answer 16

Isn't the intergal of 1/(x-1) ln|x-1) though?

Answer 17

ln|x-1|+C

Answer 18

@tkhunny @SithsAndGiggles

Answer 19

Yes it is.

Answer 20

What the... Why does Wolfram give me something different? >.< ?

Answer 21

Wolfram probably factored the denominator this way:
\[x^2-1=(1-x)(1+x)\]
If you did the same, you'd get the same answer.

Answer 22

Ah yes. Wolfram surprisingly was wrong! :P . Thanks guys :) .

Answer 23

Wolfram probably wasn't privvy to the Domain.

Answer 24

I suppose so.