What is the integral of (3x+7)/(6x^2+4x-2)

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The integral `int (3x+7)/(6x^2+4x-2) dx` has to be determined.

First find the partial fraction form of `(3x+7)/(6x^2+4x-2)` . Start with factoring the denominator.

`(3x +7)/(6x^2 + 4x - 2)`

= `(3x +7)/(6x^2 + 6x - 2x - 2)`

= `(3x +7)/(6x(x + 1) - 2(x + 1))`

= `(3x +7)/((6x - 2)(x + 1))`

= `A/(6x - 2) + B/(x + 1)`

Now,  A and B have to be determined.

`A/(6x - 2) + B/(x + 1)`

= `(A(x + 1) + B(6x - 2))/((6x - 2)(x + 1))`

`A(x + 1) + B(6x - 2) = 3x + 7`

=> `x(A + 6B) + A - 2B = 3x + 7`

A + 6B = 3 and A - 2B = 7

8B = -4

=> B = -1/2

A = 6

`int (3x+7)/(6x^2+4x-2) dx`

= `int 3/(3x - 1) - 1/2*(1/(x+1)) dx`

= `int 3/(3x - 1) dx - int (1/2)*(1/(x+1)) dx`

= `3*ln(3x - 1)/3 - ln(x+1)/2`

= `ln(3x - 1) - ln(x + 1)/2`

The integral `int (3x+7)/(6x^2+4x-2) dx = ln(3x - 1) - ln(x + 1)/2 + C`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial