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

Jan 28, 2017

$\int \frac{x \mathrm{dx}}{2 x - 1} ^ 2 = \frac{1}{4} \ln \left\mid 2 x - 1 \right\mid - \frac{1}{4} \frac{1}{2 x - 1} + C$

Explanation:

Write the numerator of the integrand function as:

$x = \frac{1}{2} \left(2 x - 1\right) + \frac{1}{2}$

So that we can split the integral as:

$\int \frac{x \mathrm{dx}}{2 x - 1} ^ 2 = \frac{1}{2} \int \frac{2 x - 1}{2 x - 1} ^ 2 \mathrm{dx} + \frac{1}{2} \int \frac{\mathrm{dx}}{2 x - 1} ^ 2$

SImplify the function in the first integral:

$\int \frac{x \mathrm{dx}}{2 x - 1} ^ 2 = \frac{1}{2} \int \frac{\mathrm{dx}}{2 x - 1} + \frac{1}{2} \int \frac{\mathrm{dx}}{2 x - 1} ^ 2$

Now note that: $d \left(2 x - 1\right) = 2 \mathrm{dx}$:

$\int \frac{x \mathrm{dx}}{2 x - 1} ^ 2 = \frac{1}{4} \int \frac{d \left(2 x - 1\right)}{2 x - 1} + \frac{1}{4} \int \frac{d \left(2 x - 1\right)}{2 x - 1} ^ 2$

and we have:

$\int \frac{x \mathrm{dx}}{2 x - 1} ^ 2 = \frac{1}{4} \ln \left\mid 2 x - 1 \right\mid - \frac{1}{4} \frac{1}{2 x - 1} + C$

Jan 28, 2017

I got: $= \frac{1}{4} \left(\ln | 2 x - 1 | - \frac{1}{2 x - 1}\right) + c$

Explanation:

We can try seting: $2 x - 1 = t$ so that:
$x = \frac{1}{2} \left(t + 1\right)$
$\mathrm{dx} = \frac{1}{2} \mathrm{dt}$
Substituting we get:
$\int \frac{x \mathrm{dx}}{2 x + 1} ^ 2 = \int \frac{1}{2} \left(t + 1\right) \frac{1}{t} ^ 2 \frac{\mathrm{dt}}{2} =$
Rearrange:
$\frac{1}{4} \int \left(\frac{t}{t} ^ 2 + \frac{1}{t} ^ 2\right) \mathrm{dt} = \frac{1}{4} \left[\int \frac{1}{t} \mathrm{dt} + \int \frac{1}{t} ^ 2 \mathrm{dt}\right] =$
and integrate:
$= \frac{1}{4} \left(\ln | t | - \frac{1}{t}\right) + c$
We finally go back to $x$ remembering that $t = 2 x - 1$:
$= \frac{1}{4} \left(\ln | 2 x - 1 | - \frac{1}{2 x - 1}\right) + c$