# Integrating ∫(\frac{4}{2x-1}): Exploring the Solution

• Physicist3
In summary, to integrate ∫( \frac{4}{2x-1} )dx, you can either factor out a 2 from the numerator and use the substitution u = 2x-1, or bring the 4 out and factor out a 2 from the denominator to make the substitution u = 2x-1. Both methods result in the integral becoming ∫( \frac{2}{u} )du, which is easily solved.

## Homework Statement

The problem that we have been given is to integrate the following: ∫( $\frac{4}{2x-1}$ )dx

## Homework Equations

I understand that the when $\frac{a}{ax+b}$ is integrated, the result is ln(ax+b) + C.

## The Attempt at a Solution

I have been told I need to make the numerator the same as the integer infront of the x term of the denominator (2) so that the equation meets the format to be $\frac{a}{ax+b}$, where a = 2 and b = -1. My lecturer informed me that I should change the 4 of the numerator from 4 to 2 and then place a 2 infront of the integral so that it looks like the following:

2∫($\frac{2}{2x-1}$).

What I can't understand is why the (2x-1) term doesn't now become (x-0.5) if the 2 at the front of the integral means that the statement will be multiplied by 2, as the top is essentially multiplied by 2 but not the bottom. I may have missed something completely obvious here but this is bugging me slightly.

Thanks

Physicist3 said:

## Homework Statement

The problem that we have been given is to integrate the following: ∫( $\frac{4}{2x-1}$ )dx

## Homework Equations

I understand that the when $\frac{a}{ax+b}$ is integrated, the result is ln(ax+b) + C.

## The Attempt at a Solution

I have been told I need to make the numerator the same as the integer infront of the x term of the denominator (2) so that the equation meets the format to be $\frac{a}{ax+b}$, where a = 2 and b = -1. My lecturer informed me that I should change the 4 of the numerator from 4 to 2 and then place a 2 infront of the integral so that it looks like the following:

2∫($\frac{2}{2x-1}$).

What I can't understand is why the (2x-1) term doesn't now become (x-0.5) if the 2 at the front of the integral means that the statement will be multiplied by 2, as the top is essentially multiplied by 2 but not the bottom. I may have missed something completely obvious here but this is bugging me slightly.

Thanks

The reason you factor out a 2 from the numerator is so that the substitution is obvious.

##u = 2x-1 \Rightarrow du = 2dx##

Another way to go is to bring the 4 in the numerator out, and factor 2 out of the denominator.
$$\int \frac{4 dx}{2x - 1} = \frac{4}{2} \int \frac{dx}{x - 1/2} = 2 \int \frac{dx}{x - 1/2}$$

The last integral is pretty easy to do.

Personally, I would not do it that way, anyway. Starting from $\int 4dx/(2x- 1)$ I would say "let u= 2x- 1. Then du= 2dx". Now I can divide both sides by 2 to get "(1/2)du= dx" and the integral becomes $\int (1/2)(4) du/u= (2)\int du/u$ which gives the same thing.