# How do you use differentiation to find a power series representation for #1/(6+x)^2#?

##### 1 Answer

Basically:

1. *Integrate to get a form of #1/(1-x)#.*

2.

*Modify the equation to achieve getting precisely*#1/(1-X)# where X is some variant of x, whether it's #x/6# (here), #-y/2# , #theta/pi# , etc.

3.

*Write out the power series with*#x# = some substituted value, like #x/6# , here.

4.

*Reverse what you did to re-acquire the original function. i.e. re-multiply by what you divided by (-1 and 1/6, here), then re-differentiate. Whatever returns your original function.*

Notice how the power series

Similarly, use -x instead of x. Every odd power is negative, and every even power is positive by virtue of squaring to some order of magnitude (e.g.

Integrate

x/6 is your new x. Plug it in, use this alternating series from a few lines up, factor in the 1/6 to get back to

Then, re-differentiate the result to get back to