integrate partial fraction

[feline]

I need to integrate 1/(2x-1)^5

I am struggling to solve the partial fraction before I can go any further

I have got

A/(2x-1)^4 + B/(2x-1)^3 + C/(2x-1)^2 + D/(2x-1) + E = 1

which I wrote as:

A(2x-1)^4 + B(2x-1)^3 + C(2x-1)^2 + D(2x-1) + E = 1

but I am struggling to get any further, I have tried substituting different values in for x but I just cant get any where with it just seem to keep going around i circles unfortuantely going no where :-(

Any help would be greatly appreciated.

Many Thanks

 

[pka]

I need to integrate 1/(2x-1)^5I am struggling to solve the partial fraction before I can go any further

Why use partial fractions?What is the derivative of \(\displaystyle \dfrac{-(2x-5)^{-4}}{8}~?\)

 

[Deleted member 4993]

I need to integrate 1/(2x-1)^5

I am struggling to solve the partial fraction before I can go any further

I have got

A/(2x-1)^4 + B/(2x-1)^3 + C/(2x-1)^2 + D/(2x-1) + E = 1

which I wrote as:

A(2x-1)^4 + B(2x-1)^3 + C(2x-1)^2 + D(2x-1) + E = 1

but I am struggling to get any further, I have tried substituting different values in for x but I just cant get any where with it just seem to keep going around i circles unfortuantely going no where :-(

Any help would be greatly appreciated.

Many Thanks

To get to the answer that pka suggested - another way would be to substitute:

2x - 5 = u

dx = du/2

 

[soroban]

Hello, feline!

Some errors . . .

First of all, use the method suggested by pka and Subhotosh.
Second, your partial fraction set-up is sloppy.
Third, have you never solved a partial fraction with repeated factors?
. . If that's true, why were you given one with five repeated factors?
Fourth, I will explain below . . .

\(\displaystyle \displaystyle\int\frac{dx}{(2x-1)^5}\)

If you must use Partial Fractions . . .

We have: .\(\displaystyle \displaystyle\frac{1}{(2x-1)^5} \;=\;\frac{A}{2x-1} + \frac{B}{(2x-1)^2} + \frac{C}{(2x-1)^3} + \frac{D}{(2x-1)^4} +\frac{E}{(2x-1)^5}\)

The obvious solution is: .\(\displaystyle \begin{Bmatrix}A = 0 \\ B=0 \\ C-0 \\ D=0 \\ E = 1\end{Bmatrix}\)

And we have: .\(\displaystyle \dfrac{1}{(2x-1)^5} \;=\;\dfrac{1}{(2x-1)^5}\) . . . . duh!

This indicates that Partial Fractions was simply a waste of time.