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

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### 1 Answer

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`