Integrate x/(6x-5) dx

The answer is 1/6 x + 5/36 ln abs (36x-30) + C... But I don't understand why it is 5/36 times ln.. I thought that when you factored out the 5 you would get 5 times ln abs (36x-30)..

I believe you are referring from your post last Thursday.

http://www.jiskha.com/display.cgi?id=1447957757

and you accept that the long division yields

x/(6x-5) = 1/6 + 5/(36x) + 25/(216x^2) + 125/(1296x^3 + ..

let's look at the term that caused your confusion

5/(36x)
= (5/36) (1/x)
integrating that give us (5/36) ln x
which was the 2nd term in my integration answer

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I think you should go with Steve's method which yields an exact integral of only 2 term plus a constant
(notice his 2nd term is not the same as my 2nd term, my expansion has an infinite number of terms. But both series are correct )

Just do one step of the long division to get

x/(6x-5) = 1/6 + (5/6)( 1/(6x-5)
now all you have to do is integrate those two terms

integral of 1/6 + (5/6)( 1/(6x-5) dx
= (1/6)x + (5/6)(1/6)ln(6x-5) ) + c
= x/6 + (5/36) ln ( 6x-5 ) + c

check by taking the derivative

and watch out for that "c"

To integrate the function x/(6x-5), we can use a technique called partial fractions.

First, let's factor the denominator: 6x - 5. This can be factored as 6(x - 5/6).

Next, we can express x/(6x - 5) as the sum of two fractions. This is done by using the partial fractions decomposition method.

x/(6x - 5) = A/(x - 5/6)

To find the value of A, we can multiply both sides of the equation by (x - 5/6):

x = A(x - 5/6)

Expanding this equation:

x = Ax - 5A/6

Now we can equate the coefficients of x on both sides:

1 = A

So, A = 1.

Now we can rewrite the integral as:

∫ (1/(x - 5/6)) dx

To integrate this, we can use the substitution method. Let u = x - 5/6, then du = dx.

Replacing x and dx in terms of u, the integral becomes:

∫ (1/u) du

This is a natural logarithmic integral. The integral of 1/u is ln|u|.

So, the final result is:

∫ (x/(6x - 5)) dx = ln|x - 5/6| + C

But remember that we initially factored out the 6 from the denominator, so instead of ln|x - 5/6|, we have ln|6x - 5|.

Therefore, the correct answer is:

∫ (x/(6x - 5)) dx = 1/6 x + 5/36 ln|6x - 5| + C

Please note that the constant of integration, C, represents an arbitrary constant and should be included in the final answer.