Question

evaluate the integral from 2 to 3: 1/(x^(2)-1)

Answer 1

\[x^2-1= (x-1)(x+1)\]\[\int\limits{\frac{dx}{x^2-1}}\]\[=\int\limits{\frac{dx}{(x-1)(x+1)}}\]
use paritial fractions to break \[\frac{1}{(x-1)(x+1)}\]
then integrate

Answer 2

\[\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}=\frac{A(x+1)+B(x-1)}{(x-1)(x+1)}\]
so this means we have
\[1=A(x+1)+B(x-1)=Ax+Bx+A-B=(A+B)x+(A-B)\]
=>
\[A+B=0\]
\[A-B=1\]

Answer 3

you can also use a hyperbolic substitution x = cosh theta

Answer 4

\[A-B=1=> A=1+B\]
\[A+B=0=> (1+B)+B=0=> 1+2B=0 => B=\frac{-1}{2}\]
\[A=1+B=> A=1+\frac{-1}{2}=\frac{1}{2}\]

Answer 5

A+B=0

A−B=1
How do u know this?

Answer 6

remember left side=right side =>
coefficient of x on left side=coefficient of x on right side
constant term on left side=constant on right side
\[\int\limits_{}^{}(\frac{1}{2} \cdot \frac{1}{x-1}+\frac{-1}{2} \cdot \frac{1}{x+1}) dx\]

Answer 7

0x+1=(A+B)x+(A-B)
how can this statement be true?

Answer 8

if what?

Answer 9

the coefficient of x on left side must equal coefficient of x on right side
the constant term on left must equal constant term on right

Answer 10

we want theses expressions to be exactly the same
0x+1
and
(A+B)x+(A-B)

Answer 11

the only way this can happen is if A+B=0 and if A-B=1

Answer 12

do you understand?
i don't think i can be anymore clear i'm sorry

Answer 13

lol yes, sorry it just took me a minute but that makes perfect sense, i see it now....so now that i have my a and b...i plug it back in, right

Answer 14

yes

Answer 15

\[\int\limits\limits_{}^{}(\frac{1}{2} \cdot \frac{1}{x-1}+\frac{-1}{2} \cdot \frac{1}{x+1}) dx \]

Answer 16

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

Answer 17

ok but we started with a definite integral

Answer 18

correct

Answer 19

\[\int\limits_{2}^{3}\frac{1}{x^2-1} dx=[\frac{1}{2}\ln|x-1|-\frac{1}{2}\ln|x+1|]_2^3\]

Answer 20

\[=(\frac{1}{2}\ln|3-1|-\frac{1}{2} \ln|3+1|)-(\frac{1}{2}\ln|2-1|-\frac{1}{2} \ln|2+1|)\]

Answer 21

\[=\frac{1}{2}\ln|2|-\frac{1}{2}\ln|4|-\frac{1}{2}\ln|1|+\frac{1}{2}\ln|3|\]

Answer 22

\[=\frac{1}{2}\ln(2)-\frac{1}{2}\ln(4)-0+\frac{1}{2}\ln(3)\]

Answer 23

\[=\ln(\sqrt{2})-\ln(\sqrt{4})+\ln(\sqrt{3})=\ln(\sqrt{2})-\ln(2)+\ln(\sqrt{3})=\ln(\frac{\sqrt{2} \sqrt{3}}{2})\]
\[=\ln(\frac{\sqrt{6}}{2})\]

Answer 24

thank u! I understand this one now haha

Answer 25

i'm happy for you :)

Answer 26

i wish i could fan myinanaya twice and give her medal 5 times for every question :)

Answer 27

method using x = cosh theta:

let x = cosh theta
dx/d theta = sinh theta
\[\frac{1}{x^2-1} = \frac{1}{\cosh^2 \theta - 1}=\frac{1}{\sinh^2 \theta}\]
Therefore
\begin{eqnarray*}
\int{\frac{1}{x^2-1}}dx&=&\int{\frac{1}{\sinh^2 \theta}}\sinh \theta d\theta \\
&=&\int{\cosech \theta}d\theta \\
&=&-\coth \theta\cosech \theta + C
\end{eqnarray*}
Since \[\theta = \cosh^{-1} x\]
answer is
\[\coth (\cosh^{-1} x) \cosech (\cosh^{-1} x) + C\]

Answer 28

eww hyperbolic trig functions

Answer 29

forgot the minus sign at the very end

Answer 30

good job xactxx

Answer 31

result should be the same since inverse hyperbolic functions are in terms of ln as well ;)

Answer 32

yes i would check but i don't want too lol

Answer 33

i believe you

Answer 34

let me see

\[-\coth(\cosh^{−1}3)\cosech(\cosh^{−1}3) +\coth(\cosh^{−1}2)\cosech(\cosh^{−1}2)\]

cosh^-1 3 = 1.763, cosh^-1 2 = 1.317.

so

\[-\coth 1.763 \cosech 1.763 + \coth 1.317 \cosech 1.317\]
\[=-1.061 x 0.353 + 1.732 x 0.577\]
\[=-0.665 + 1\]
\[=0.334\]

Possibly correct answer is 1/3?

Answer 35

http://www.wolframalpha.com/input/?i=integrate%281%2F%28x^2-1%29%2C+x%3D2..3%29

Answer 36

http://www.wolframalpha.com/input/?i=integrate%281%2F%28x^2-1%29%29

Answer 37

they are saying the antiderivative is negative tan hyperbolic inverse+C