# Thread: Integration of f(x)^n Where n is -ve fraction

1. ## Integration of f(x)^n Where n is -ve fraction

I am trying to integrate the following problem with limit +D and -D

$\int{\frac{1}{(z^2 + r^2)^\frac{3}{2}}\,dz}$

Now intuition tells me to treat it like

$\int{(z^2 + r^2)^\frac{-3}{2}\,dz}$

Here is where I'm going wrong - the way I'm integrating gives me

$\frac{(2z)(z^2 + r^2)^\frac{-1}{2}}{\frac{-1}{2}}$

$\frac{-4z}{(z^2 + r^2)^\frac{-1}{2}}$

But if I differentiate that I get

$\frac{4r^2}{(z^2 + r^2)^\frac{3}{2}}$

So have randomly created a factor of $4r^2$. Therefore I'm doing something wrong but I can't seem to get a rule to fit the situation.

Differentiating

$\int{\frac{z}{r^2(z^2 + r^2)^\frac{1}{2}}\,dz}$

Returns me to the original problem

$\frac{1}{(z^2 + r^2)^\frac{3}{2}}$

But that would imply that the basic integral is divided by $r^2$ despite the fact that for all intents and purposes $r^2$ is a constant.

Can anyone explain, as I think I've horribly confused myself about something not that complex?

2. Originally Posted by M1ke
I am trying to integrate the following problem with limit +D and -D

$\frac{1}{(z^2 + r^2)^\frac{3}{2}}$

Now intuition tells me to treat it like

$(z^2 + r^2)^\frac{-3}{2}$

Here is where I'm going wrong - the way I'm integrating gives me

$\frac{(2z)(z^2 + r^2)^\frac{-1}{2}}{\frac{-1}{2}}$

$\frac{-4z}{(z^2 + r^2)^\frac{-1}{2}}$

But if I differentiate that I get

$\frac{4r^2}{(z^2 + r^2)^\frac{3}{2}}$

So have randomly created a factor of $4r^2$. Therefore I'm doing something wrong but I can't seem to get a rule to fit the situation.

Differentiating

$\frac{z}{r^2(z^2 + r^2)^\frac{1}{2}}$

Returns me to the original problem

$\frac{1}{(z^2 + r^2)^\frac{3}{2}}$

But that would imply that the basic integral is divided by $r^2$ despite the fact that for all intents and purposes $r^2$ is a constant.

Can anyone explain, as I think I've horribly confused myself about something not that complex?
Which variable are you integrating with respect to?

Is it $\int{\frac{1}{(z^2 + r^2)^\frac{3}{2}}\,dz}$ or $\int{\frac{1}{(z^2 + r^2)^\frac{3}{2}}\,dr}$?

3. sorry - all by $dz$.

Will amend the original post.

4. Try $z = r \cdot \sinh \left( t \right)$, $dz = r \cdot \cosh \left( t \right)dt$.

5. Try to use the hyperbolic functions.

Then

$\int {\frac{{dz}}{{{{\left( {{z^2} + {r^2}} \right)}^{{3 \mathord{\left/{\vphantom {3 2}} \right.\kern-\nulldelimiterspace} 2}}}}}} = \left\{ \begin{gathered}z = r \cdot \sinh \left( t \right), \hfill \\dz = r \cdot \cosh \left( t \right)dt \hfill \\ \end{gathered} \right\} = \frac{1}{{{r^2}}}\int {\frac{{dt}}{{{{\cosh }^2}\left( t \right)}}} = \frac{1}{{{r^2}}}\tanh \left( t \right) + C =$

$= \frac{1}{{{r^2}}}\frac{{\sinh \left( t \right)}}
{{\cosh \left( t \right)}} + C = \frac{1}
{{{r^2}}}\frac{{\sinh \left( {{\text{arcsinh}}\left( {{z \mathord{\left/
{\vphantom {z r}} \right.\kern-\nulldelimiterspace} r}} \right)} \right)}}
{{\cosh \left( {{\text{arcsinh}}\left( {{z \mathord{\left/
{\vphantom {z r}} \right.\kern-\nulldelimiterspace} r}} \right)} \right)}} + C =$
$\frac{1}{{{r^2}}}\frac{{{z \mathord{\left/
{\vphantom {z r}} \right.\kern-\nulldelimiterspace} r}}}
{{\sqrt {1 + {{{z^2}} \mathord{\left/{\vphantom {{{z^2}} {{r^2}}}} \right.\kern-\nulldelimiterspace} {{r^2}}}} }} + C = \frac{z}
{{{r^2}\sqrt {{z^2} + {r^2}} }} + C.$

6. People tend to be more familar with the trig functions so I might recommend

$z = r \tan \theta$

BTW - it reduces to a nice integral afterwards.