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.