# `int (x^2 + 1)/((x - 3)(x - 2)^2) dx` Evaluate the integral

Integrate `int(x^2+1)/[(x-3)(x-2)^2]dx`

Rewrite the rational function using partial fractions.

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

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

`x^2+1=A(x^2-4x+4)+B(x^2-5x+6)+Cx-3C`

`x^2+1=Ax^2-4Ax+4A+Bx^2-5Bx+6B+Cx-3C`

`x^2+1=(A+B)x^2+(-4A-5B+C)x+(4A+6B-3C)`

Equate coefficients and solve for A, B, and C.

`1=A+B`                     (1)

`0=-4A-5B+C`     (2)

`1=4A+6B-3C`       (3)

Adding Equations (2) and (3) will give you

`1=B-2C`

`B=1+2C`    (4)

Using equation (1) substitute variable A with  `A=1-B`

into equation (3).

Using equation (4) substitute variable B with `B=1+2C`

equation (3).

`1=4A+6B-3C`     (3)

`1=4(1-B)+6(1+2C)-3C`

`1=4-4B+6+12C-3C`

`1=10-4B+9C`

`-9=-4(1+2C)+9C`

`-9=-4-8C+9C`

`-5=C`

`B=1+2C`

`B=1+2(-5)`

`B=1-10`

`B=-9`

`A=1-B`

`A=1-(-9)`

`A=10`

`int(x^2+1)/[(x-3)(x-2)^2]dx=int10/(x-3)dx+int-9/(x-2)dx+int-5/(x-2)^2dx`

`=10ln|x-3|-9ln|x-2|+5/(x-2)+C`

`=10ln|x-3|-9ln|x-2|+5/(x-2)+C `

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You need to use partial fraction decomposition, such that:

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

`x^2+1 = a(x-2)^2 + b(x-2)(x-3) + c(x-3)`

`x^2+1 = ax^2 - 4ax + 4a + bx^2 - 5bx + 6b + cx - 3c`

Group the terms:

`x^2+1 = x^2(a+b) + x(-4a-5b+c) + 4a + 6b - 3c`

`a + b = 1 => a = 1-b`

`-4a-5b+c = 0 => c = 4a+5b`

`6b - 3c = 1 => 6b - 3(4a+5b) = 1 => -9b - 12a = 1`

`-9b - 12(1-b) = 1 => 3b - 12 = 1 => 3b = 13 => b = 13/3`

`a = -10/3`

`c = -40/3 + 65/3 => c = 25/3`

`(x^2+1)/((x-3)(x-2)^2) = -10/(3(x-3)) + 13/(3(x-2)) + 25/(3(x-2)^2)`

Taking integral both sides yields:

`int (x^2+1)/((x-3)(x-2)^2) dx = int -10/(3(x-3)) dx + int 13/(3(x-2)) dx + int 25/(3(x-2)^2)dx`

`int (x^2+1)/((x-3)(x-2)^2) dx = -10/3*ln|x-3| + 13/3ln|x-2| - 25/(3(x-2)) + c`

Hence, evaluating the give integral yields  `int (x^2+1)/((x-3)(x-2)^2) dx = -10/3*ln|x-3| + 13/3ln|x-2| - 25/(3(x-2)) + c.`

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