Question

I'm not sure what I'm doing wrong here, just a simple error somewhere, but I'm trying to show by partial fraction decomposition that tanh^-1(x) = 1/2ln((1+x)/(1-x)), but I'm running into some trouble and somehow the arguments for my natural logarithms are the negative of what they should be. My steps posted below in a few minutes.

Answer 1

\[\frac{ d }{ dx }\tanh ^{-1}(x) = \frac{ 1 }{ 1-x ^{2} };\]
\[\frac{ 1 }{ 1-x ^{2} }=\frac{ A }{ (1+x) }+\frac{ B }{ (1-x) }.\]
\[\frac{ 1(1+x)(1-x) }{ (1+x)(1-x) } = \frac{ A(1+x)(1-x) }{ (1+x) }+\frac{ B(1-x)(1+x) }{ (1-x) }\]

Answer 2

\[1 = A(1-x)+B(1+x)\]
Since this is an identity, all values of x are valid; I'll assume x = 1. \[1 = A(1-1)+B(1+1);\]\[1 = 2B; \]\[B = \frac{ 1 }{ 2 }\]
Doing the same for x where x = -1 results in A = -1/2. (cont'd)

Answer 3

\[\int\limits_{}^{}\frac{ (-1/2) }{ (1-x) }+ \int\limits_{}^{}\frac{ (1/2) }{ (1+x) } = -\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ 1-x }+\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ 1+x }=\]

Answer 4

\[-\frac{ 1 }{ 2 }\ln(1-x)+\frac{ 1 }{ 2 }\ln(1+x) = \frac{ 1 }{ 2 }\ln \frac{ (1+x) }{ (1-x) }\]
Wait, what? I think I just did it right, rofl.

Answer 5

Nope, I didn't, I switched the denominators of A and B.

Answer 6

But somewhere in there, if the denominators are correct, I'm ending up with the wrong result where the argument of the log is the reciprocal of what it should be.

Answer 7

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

Answer 8

what is the derivative of inverse tan hyperbolic function ?

Answer 9

At the top, 1/(1-x^2).

Answer 10

oh, k ...
and the condition for that is |x| <1

Answer 11

Yeah, on that condition. Unless I'm losing my marbles,

Answer 12

I'm messing up hardcore somewhere, I just can't see where

Answer 13

during integral of 1/1-x
you should also divide by the co-efficient of x, that is -1

Answer 14

I don't know what you mean when you mentioned dividing by the coefficient of x during the integral, what?

Answer 15

int (1/2)/(1-x) dx = 1/2 (ln (1-x)) * (-1)

Answer 16

like \(\int \dfrac{1}{ax+b}dx =(1/a)\ln |ax+b|+c\)

Answer 17

because d/dx (ln[ax+b]) = a/ax+b

Answer 18

Oh! Derp. Okay, wait just one second.

Answer 19

and A comes out to be +1/2 -_-

Answer 20

1= A (1- (-1))
1=A(1+1)

Answer 21

I dun goof'd.

Answer 22

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

Answer 23

(Cont'd for logarithm/integration part)

Answer 24

its correct, now u will get it

Answer 25

\[\int\limits_{}^{}\frac{ (1/2) }{ (1+x) }dx+\int\limits_{}^{}\frac{ (1/2) }{ (1-x) }dx = \frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ 1+x }+\frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ 1-x }\]\[\frac{ 1 }{ 2 }\ln(1+x)+\frac{ 1 }{ 2 }(-1)\ln(1-x) = \frac{ 1 }{ 2 }\ln \frac{ (1+x) }{ (1-x) }\]

Answer 26

Huzzah.

Answer 27

\(\huge \checkmark\)

Answer 28

How did you do that? (check)

Answer 29

I know you can use LaTeX to write equations here, was it that, or something else?

Answer 30

latex only
`\(\checkmark\)`

Answer 31

try to give it a color too

Answer 32

\(\huge \color{red} \checkmark\)
\(\huge \color{green} \checkmark\)