how to solve integral of 3x-3/(3x^2-6x+1)?

Question

i have a problem solving this. Help?

Answer 1

Laras,

        A very simple way to do this problem is to let u=3x²-6x=1 then du=6(x-)dx and the integrand becomes (1/2)du/u which integrates to ln u + constant or ln(3x²-6x+1) + constant.

 

 

Jim

Answer 2

This is an interesting problem. You'll have to do some algebraic manipulation before taking the integral.

 

The first step is to rewrite the second degree polynomial in the denominator in the form of:

 

(x - a)² +d Hence, 3x²-6x+1 = 3(x²-2x+1)+1-3 = 3*(x-1)²-2 = 3*((x-1)² - 2/3)

 

Now, the last term could be factored further in which is:

 

3*((x-1)-√(2/3))*((x-1)+√(2/3))

 

Next the function takes on the form of:

 

3x-3/(3*((x-1)-√(2/3))*((x-1)+√(2/3)) = x-1/((x-1)-√(2/3))*((x-1)+√(2/3))

 

Next, let u = x-1 to simply the integration process (makes du=dx) in which the function becomes:

 

u/((u-√(2/3))*(u+√(2/3)))

 

In order to further simplify the integration is to rewrite the function as:

 

u/((u-√(2/3))*(u+√(2/3))) = K₁/(u-√(2/3)) + K₂/(u+√(2/3))

 

Now, you could take the integral of each term and get the answer in the form of:

 

K₁*ln(u-√(2/3))+K₂*ln(u+√(2/3)). Substitute u being x-1 into the last equation to get the final answer.

 

I'll let you take it from here to get the complete answer.