Question

Proof. Integrals and ln.

Answer 1

show \[\int\limits_{5/4}^{2}\frac{ 1 }{ 1+x^2 }dx=\frac{ 1 }{ 2 }\ln(\frac{ 1 }{ 3 })\]

Answer 2

You mean prove?

Answer 3

\[\frac{ d }{ dx }\coth^{-1}(x)=\frac{ 1 }{ 1+x^2 }\] so I guess we can do
\[(\coth^{-1}(x))_{5/4}^2\]

Answer 4

Yeah

Answer 5

\[(\coth^{-1}(2))-(\coth^{-1}(5/4))\]

Answer 6

is there a mistake in the problem

Answer 7

..where? in the original question?

Answer 8

yes

Answer 9

I'm getting one side is positive while the other side is negative

Answer 10

No there is no mistake xD
but (1/2)ln(1/3) is equivalent to (-1/2)ln(3)
if that's what you're getting

Answer 11

the equation is false

Answer 12

huh o.o
Perhaps I copied it down wrong from the white board...let me check with a classmate

Answer 13

I think there should be a minus sign between 1 and x^2 ?

Answer 14

aah yeah it's a minus. sorry.

Answer 15

do you want to do this the fun way
or do you want to do this the normal way that everyone else might do it kind away

Answer 16

haha lets do it the fun way! as long as it's not too abstract o.o

Answer 17

ok do you know the definition of ln(x)?

Answer 18

(I think i'm meant to use the cosh stuff though..)
and yes.

Answer 19

\[\ln(t)=\int\limits_1^t \frac{1}{x} dx \\ \ln(3^\frac{-1}{2})=\int\limits_1^{3^{\frac{-1}{2}}} \frac{1}{x} dx\]

Answer 20

so we want to show the following:
\[\int\limits_\frac{5}{4}^2 \frac{1}{1-x^2} dx=\int\limits_1^{3^\frac{-1}{2}} \frac{1}{x} dx\]

Answer 21

i though there is some way to do this without integrating
one sec while I think

Answer 22

If you want to integrate, partial fractions work nicelyh :

\[\dfrac{1}{1-x^2} = \dfrac{1}{2}\left(\dfrac{1}{1+x}+\dfrac{1}{1-x}\right)\]

Answer 23

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

Answer 24

err you would think there would be some substitution you can do to show those two integrals are equal

Answer 25

= \[\frac{ 1 }{ 2 }(\ln(1+x)\color{red}{-}\ln(1-x))\] then?

This also equals \(\text{arccoth} x\)

Answer 26

huh. that whole thing equals arccoth(x)?

Answer 27

Can I just take that and evaluate it at the 2 and 5/4?

Answer 28

http://mathworld.wolfram.com/InverseHyperbolicCotangent.html

Answer 29

\[\text{arccoth}x = \dfrac{1}{2}\left(\ln(x+1)-\ln(x-1)\right)\]

Answer 30

\[\coth^{-1}(x)\] evaluated at 2 and 5/4
right? but that's the part I get stck at

Answer 31

We don't really need to use coth^-x here

Answer 32

just evaluate the bounds on logs

Answer 33

\[\dfrac{1}{2}\left(\ln(x+1)-\ln(x-1)\right)\]

evaluate above expression at x = 2 and 5/4

Answer 34

\[\frac{ 1 }{ 2 }[(\ln(1+2)-\ln(1-2))-(\ln(1+\frac{ 5 }{ 4 })-\ln(1-\frac{ 5 }{ 4 })]\]

Answer 35

just double check the sign..

Answer 36

It's all correcct, right?

Answer 37

\[\frac{ 1 }{ 2 }[\ln(\frac{ 3 }{ -1 })-\ln (\frac{ 9/4 }{ -1/4 })]\]

Answer 38

\[\frac{ 1 }{ 2}[\ln(-3)-\ln(9)]\]

Answer 39

\[\frac{ 1 }{ 2 }\ln(- \frac{ 1 }{ 3 })\] errr I got a negative in there hm.

Answer 40

oh I got it :) it's -9 not 9
the ln((9/4)/(-1/4))

Answer 41

maybe either way you have to go you have to do partial fractions
\[\text{so \let's look at this again } \\ \text{ we have } \\ -\frac{1}{2} \int\limits _\frac{5}{4}^2 (\frac{-1}{1+x}+\frac{-1}{1-x}) dx=-\frac{1}{2} \int\limits _1^{3} \frac{1}{x} dx \\ \\ \\ \text{ since } -\frac{1}{2} \ln(3)=\frac{-1}{2} \int\limits_1^{3} \frac{1}{x} dx \\ \text{ so we have \to prove } \int\limits_\frac{5}{4}^{2} (\frac{-1}{x+1}+\frac{-1}{1-x})dx =\int\limits_1^3 \frac{1}{x} dx \\ \int\limits_\frac{5}{4} ^2 \frac{-1}{x+1} dx+\int\limits_\frac{5}{4}^2 \frac{-1}{1-x} dx =\int_1^3 \frac{1}{x} dx \]
now we can do a sub
first integral do u=-(x+1)
du=-dx

second sub do v=-(1-x)
dv=dx

so we have
\[\int\limits_\frac{-9}{4}^{-3} \frac{1}{u} (-du)+\int\limits_{\frac{1}{4}}^{1} \frac{1}{v} dv \\ \]
\[\text{ \let } u=-g \\ du=-dg \\ \int\limits_\frac{9}{4}^{3} \frac{1}{-g} dg +\int\limits_\frac{1}{4}^1 \frac{1}{v} dv \\ -\int\limits_\frac{9}{4}^3 \frac{1}{g} dg +\int\limits_\frac{1}{4}^ 1 \frac{1}{v} dv \\ \\ \]

\[-(\int\limits_\frac{9}{4}^1 \frac{1}{g} dg +\int\limits_1^3 \frac{1}{g} dg)-\int\limits_{1}^\frac{1}{4} \frac{1}{v} dv \\ \int\limits_1^\frac{9}{4} \frac{1}{g} dg -\int\limits_1^3 \frac{1}{g} dg-\int\limits_1^\frac{1}{4} \frac{1}{v} dv \\ \text{ then by definition of \log... } \\ \ln(\frac{9}{4})-\ln(3)-\ln(\frac{1}{4}) \\ \ln(9)-\ln(4)-\ln(3)-\ln(1)+\ln(4) \\ \ln(9)-\ln(3) \\ \ln(3^2)-\ln(3) \\ 2 \ln(3)-\ln(3) \\ \ln(3) \]\
which is the other side
lol maybe this way is too fun :p

Answer 42

ok let me check out what you have done

Answer 43

oh my xD

Answer 44

I had to play with it but i did manage to work around the integrating part

Answer 45

there is probably one thing I would change
and I think it was a major problem for you
when you integrate 1/x put ln|x| instead of ln(x)
the negatives won't give you misery now

Answer 46

Ooh thanks :D that will relieve me of much stress haha
can you help me with the steps from
\[\frac{ 1 }{ 1-x^2 }=\frac{ 1 }{ 2 }(\frac{ 1 }{ 1+x }-\frac{ 1 }{ 1-x })\] ? how to I get from part a to part b?

Answer 47

\[\frac{ 1 }{ 2 }[(\ln|1+2|-\ln|1-2|)-(\ln|1+\frac{ 5 }{ 4 }|-\ln|1-\frac{ 5 }{ 4 }|] \\ \frac{1}{2} (\ln(3)-\ln(1))-(\ln(\frac{9}{4})-\ln(\frac{1}{4}))\]

Answer 48

ln|-1| is just ln(1)
and that ln|-1/4| is just ln(1/4)
yes...

Answer 49

\[\frac{1}{1-x^2}=\frac{1}{(1-x)(1+x)}=\frac{A}{1-x}+\frac{B}{1+x}\]

Answer 50

\[\frac{1}{1-x^2}=\frac{A}{1-x}+\frac{B}{1+x} \\ \frac{1}{1-x^2}=\frac{A(1+x)+B(1-x)}{(1-x)(1+x)} \\ \frac{1}{1-x^2}=\frac{A(1+x)+B(1-x)}{1-x^2}\]

Answer 51

implies 1=A(1+x)+B(1-x)

Answer 52

you solve for A and B such that both sides are the same

Answer 53

you could use heaviside method if you want

Answer 54

if both sides are the same then no matter what x you choose the equality should hold

for example of we choose x=-1
1=A(1+(-1))+B(1-(-1))
1=A(0)+B(-2)
1=B(-2)

solve for B
you can find A in a similar way

Answer 55

ah.
What do a and b represent though? in that whole process o.o

Answer 56

The constant values that we want to find

Answer 57

you haven't done partial fractions?

Answer 58

and where do we pull the 1/2 from then

Answer 59

well if you solve that one equation for B you get B is -1/2

Answer 60

oops B is 1/2 I made a mistake above

Answer 61

1=A(1+(-1))+B(1-(-1))
1=A(0)+B(2)
1=B(2)
B=1/2

Answer 62

you should also find A=1/2 in a similar way

Answer 63

then you will get
\[\frac{1}{1-x^2}=\frac{\frac{1}{2}}{1+x}+\frac{\frac{1}{2}}{1-x}\]
I replaced A and B with 1/2

Answer 64

ok the first step in doing partial fractions
make sure the poly on top is less in degree than the poly on bottom

Answer 65

we have that

Answer 66

then we factor 1-x^2

Answer 67

(1+x)(1-x)

Answer 68

both of those factors are linear (and we have no repeated linears)

Answer 69

so 1/(1-x^2)=A/(1-x)+B/(1+x)

we want the fractions to be proper fractions
so the numerator is one degree less than the denominator

Answer 70

we have linears on bottom so the tops are constants

Answer 71

then what I did next was combine the fractions on the right

Answer 72

the bottoms were the same
so all I had to do was compare the tops

Answer 73

o.o woah. I did not know there was so much to all this o.0

Answer 74

http://www.purplemath.com/modules/partfrac3.htm
http://www.purplemath.com/modules/partfrac.htm

Answer 75

if you want more examples

Answer 76

anyways I'm really really sorry but I'm about to be logged off

Answer 77

ok:) thank you so much for the help!!