# Simple Integration

Started by March 3, 2018
```I have an integral

1/(a^2-x^2)  wrt x.  If you use partial fractions we get

(1/2a) * ln  [ (x+a)/(x-a) ]   +c

and zero initial conditions gives c=0

however, this is a standard integral and should be

(1/a)arctanh(x/a)

How to get to that stage. I know of course arctanh(x)=(1+x)/(1-x)

when I use this the sign is wrong.
```
```On 04.03.18 02.13, gyansorova@gmail.com wrote:
> I have an integral
>
> 1/(a^2-x^2)  wrt x.  If you use partial fractions we get
>
> (1/2a) * ln  [ (x+a)/(x-a) ]   +c
>
> however, this is a standard integral and should be
>
> (1/a)arctanh(x/a)
>
> How to get to that stage. I know of course arctanh(x)=(1+x)/(1-x)

The latter is wrong.

arctanh(x) := ln((1+x)/(1-x))/2

And this will make the above terms equivalent.

Marcel

```
```On 04.03.2018 4:13, gyansorova@gmail.com wrote:
> I have an integral
>
> 1/(a^2-x^2)  wrt x.  If you use partial fractions we get
>
> (1/2a) * ln  [ (x+a)/(x-a) ]   +c
>
> and zero initial conditions gives c=0
>
> however, this is a standard integral and should be
>
>
> (1/a)arctanh(x/a)
>
> How to get to that stage. I know of course arctanh(x)=(1+x)/(1-x)
>
> when I use this the sign is wrong.
>

Gyansorova, the integral 1/(a^2-x^2)  wrt x is only defined for regions
(1) x < -a, (2) -a < x < a, and (3) x > a.

You cannot include points x = -a and x = a within the integration
limits, because that would be improper integral. Without involving
complex calculus, the best option is to use Cauchy principal value as a
common workaround for dealing with improper integrals.

However, as long as we are going to stay with real and proper integrals,
you are limited to either of the regions -- (1), (2) or (3).

Now, remember that the indefinite integral of 1/x is not ln(x) + C, as
many people seem to mistakenly assume, but ln(|x|) + C, where |x|
denotes the absolute value of x.

Which means that if you use partial fractions, for regions (1) and (3)
you get (1/2a) * ln [ (x+a)/(x-a) ] + C. While for region (2) you get
(1/2a) * ln [ - (x+a)/(x-a) ] + C.

In the same time, arctanh(x) as a real function is defined for -1<x<1 --
because tanh(x) belongs to the region (-1; 1) for any real x. Which
means that the expression (1/a) arctanh(x/a) + C is only defined for
region (2), where it coincides with the answer you get using partial
fractions.

Hope that helps!

Gene

```
```On Sunday, March 4, 2018 at 11:54:54 PM UTC+13, Evgeny Filatov wrote:
> On 04.03.2018 4:13, gyansorova@gmail.com wrote:
> > I have an integral
> >
> > 1/(a^2-x^2)  wrt x.  If you use partial fractions we get
> >
> > (1/2a) * ln  [ (x+a)/(x-a) ]   +c
> >
> > and zero initial conditions gives c=0
> >
> > however, this is a standard integral and should be
> >
> >
> > (1/a)arctanh(x/a)
> >
> > How to get to that stage. I know of course arctanh(x)=(1+x)/(1-x)
> >
> > when I use this the sign is wrong.
> >
>
> Gyansorova, the integral 1/(a^2-x^2)  wrt x is only defined for regions
> (1) x < -a, (2) -a < x < a, and (3) x > a.
>
> You cannot include points x = -a and x = a within the integration
> limits, because that would be improper integral. Without involving
> complex calculus, the best option is to use Cauchy principal value as a
> common workaround for dealing with improper integrals.
>
> However, as long as we are going to stay with real and proper integrals,
> you are limited to either of the regions -- (1), (2) or (3).
>
> Now, remember that the indefinite integral of 1/x is not ln(x) + C, as
> many people seem to mistakenly assume, but ln(|x|) + C, where |x|
> denotes the absolute value of x.
>
> Which means that if you use partial fractions, for regions (1) and (3)
> you get (1/2a) * ln [ (x+a)/(x-a) ] + C. While for region (2) you get
> (1/2a) * ln [ - (x+a)/(x-a) ] + C.
>
> In the same time, arctanh(x) as a real function is defined for -1<x<1 --
> because tanh(x) belongs to the region (-1; 1) for any real x. Which
> means that the expression (1/a) arctanh(x/a) + C is only defined for
> region (2), where it coincides with the answer you get using partial
> fractions.
>
> Hope that helps!
>
> Gene

It does thanks. Sorry about the typo in the question, should have been a ln there as
previous poster pointed out
```