# Determine the integral [(3x^2+2x+1)/(x+1)(x^2-1)]dxi've tried to use partial fractions on this, but I'm not sure if my answer is correct. My answer: 3ln(x+1) + 2ln(x^2-1) + C

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You may use partial fraction decomposition such that:

`(3x^2+2x+1)/(x+1)(x^2-1) = A/(x+1) + B/(x+1)^2 + C/(x-1)`

You need to bring the fraction to the right to a common denominator such that:

` 3x^2+2x+1 = A(x^2-1) + B(x-1) + C(x+1)^2 `

` 3x^2+2x+1 = Ax^2 - A + Bx - B + Cx^2 + 2Cx + C`

` A+C=3 =gt A = 3-C B+2C=2 =gt B = 2-2C -A-B+C=1 =gt C-3+2C-2+C=1 4C-5=1=gt4C=6 =gt C=6/4 C=3/2 A=3 - 3/2 =gt A=3/2 B = 2 - 3 = -1`

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

You need to integrate both sides such that:

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

`int (3x^2+2x+1)/(x+1)(x^2-1) dx = (3/2)ln|x+1| + 1/(x+1) + (3/2)ln|x-1| + c`

`int (3x^2+2x+1)/(x+1)(x^2-1) dx = (3/2)ln(x^2-1) + 1/(x+1) + c`

`int (3x^2+2x+1)/(x+1)(x^2-1) dx = ln sqrt((x^2-1)^3) + 1/(x+1) + c`

**Hence, evaluating the integral using partial fraction decomposition yields `int (3x^2+2x+1)/(x+1)(x^2-1) dx = ln sqrt((x^2-1)^3) + 1/(x+1) + c.` **

The denominator of fraction is `(x+1)(x^2-1) ` and the integral is:

`int (3x^2+2x+1)/((x+1)(x^2-1))dx = ln sqrt((x^2-1)^3) + 1/(x+1) + c`