How would you intergrate 2/(2x-1)dx?

Question

Thanks for your help.

Answer 1

Well, you have to recognize a special case about this integral. You may notice that numerator is the derivative of the denominator in this equation, and this mirrors one of the cases that you should already know. Specifically,

∫[f'(x)/f(x)]dx = ln| f(x) | + C

Therefore, we see that for this case, f(x) = 2x - 1 and f'(x) = 2.

Therefore ∫2/(2x-1)dx = ln|2x-1| + C.

Answer 2

Lets use INT for integrate and LN for natural logarithm. The derivative of LN (x) is (1 / x) dx.

Given INT 2 / (2x - 1) dx .

Let u = 2x - 1 and du = 2 dx.

By u-substitution, the integral can be written as

INT (1 / u) du

= LN (u) + c

and since u = 2x + 1 we can write

INT 2 / (2x + 1) dx = LN (2x + 1) + c

It is possible for x to be a negative number for which LN (2x -1) is undefined. So, we will place the absolute value symbol around the argument 2x - 1.

That is, INT 2 / (2x + 1) dx = LN |2x + 1| + c

Answer 3

Use a simple substitution like z = 2x-1 and thus dz = 2 dx.

You'll see then the question is how to integrate 1/z which is easy. But don't forget to substitute back for x.

You end up with ln(2x-1) + C, where C is an arbitrary constant.

Answer 4

Here we use the u-substitution method from calculus.

Let u = 2x -1, then du = 2 dx.

So now we are integrating du/u

The antiderivative of 1/u is ln(u)

Therefore the integral would be ln|2x-1| + C.

ln|2x-1| + C is the answer

Answer 5

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Answer 6

any time you see an integral in the form of u'/u, the integral is ln |u|. So this equals ln |2x-1| + C

Answer 7

∫ 2 / (2x - 1).dx = log (2x - 1) + C

Answer 8

= ln ( 2x - 1 ) + c