`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).

...

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`

The final answer is:

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