Question

How do you evaluate the integral `int (xdx)/(2x-1)^2`?

Answer 1

`int (xdx)/(2x-1)^2 = 1/4 ln abs (2x-1) -1/4 1/(2x-1) +C`

Explanation:

Write the numerator of the integrand function as:

`x = 1/2 (2x-1) +1/2`

So that we can split the integral as:

`int (xdx)/(2x-1)^2 = 1/2 int (2x-1)/(2x-1)^2 dx +1/2 int (dx)/(2x-1)^2`

SImplify the function in the first integral:

`int (xdx)/(2x-1)^2 = 1/2 int (dx)/(2x-1) +1/2 int (dx)/(2x-1)^2`

Now note that: `d(2x-1) = 2dx`:

`int (xdx)/(2x-1)^2 = 1/4 int (d(2x-1))/(2x-1) +1/4 int (d(2x-1))/(2x-1)^2`

and we have:

`int (xdx)/(2x-1)^2 = 1/4 ln abs (2x-1) -1/4 1/(2x-1) +C`

Answer 2

I got: `=1/4(ln|2x-1|-1/(2x-1))+c`

Explanation:

We can try seting: `2x-1=t` so that:
`x=1/2(t+1)`
`dx=1/2dt`
Substituting we get:
`int(xdx)/(2x+1)^2=int1/2(t+1)1/t^2dt/2=`
Rearrange:
`1/4int(t/t^2+1/t^2)dt=1/4[int1/tdt+int1/t^2dt]=`
and integrate:
`=1/4(ln|t|-1/t)+c`
We finally go back to `x` remembering that `t=2x-1`:
`=1/4(ln|2x-1|-1/(2x-1))+c`