You need to notice that the degree of numerator is smaller than the degree of denominator, hence, you may use the partial fraction decomposition, such that:

`(5x^2-7x+22)/((x-1)(x^2+9)) = A/(x-1) + (Bx + C)/(x^2+9)`

`5x^2-7x+22 = Ax^2 + 9A + Bx^2 + Cx - Bx - C`

`5x^2-7x+22 = x^2(A+B) +...

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You need to notice that the degree of numerator is smaller than the degree of denominator, hence, you may use the partial fraction decomposition, such that:

`(5x^2-7x+22)/((x-1)(x^2+9)) = A/(x-1) + (Bx + C)/(x^2+9)`

`5x^2-7x+22 = Ax^2 + 9A + Bx^2 + Cx - Bx - C`

`5x^2-7x+22 = x^2(A+B) + x(C-B) + 9A - C`

Equating the coefficients of like powers yields:

`A+B = 5 => A = 5-B`

`C-B = -7 => B-C = 7`

`9A-C = 22 => 9(5-B) - C = 22 => 45 - 9B - C = 22 => -9B - C = -23 => 9B + C = 23`

`B - C + 9B + C = 7 + 23 => 10B = 30 => B = 3 => A = 2 => C = -4`

`(5x^2-7x+22)/((x-1)(x^2+9)) = 2/(x-1) + (3x-4)/(x^2+9)`

Integrating both sides yields:

`int (5x^2-7x+22)/((x-1)(x^2+9)) dx= 2int 1/(x-1)dx + int (3x-4)/(x^2+9)dx`

`int (5x^2-7x+22)/((x-1)(x^2+9)) dx = 2 ln|x- 1| + 3 int x/(x^2+9)dx - 4 int 1/(x^2+9)dx`

You should use the following substitution to evaluate `int x/(x^2+9)dx` such that:

`x^2 + 9 = t => 2xdx = dt => xdx = (dt)/2`

`int x/(x^2+9)dx = (1/2) int (dt)/t = (1/2) ln |t| + c`

`int (5x^2-7x+22)/((x-1)(x^2+9)) dx = 2 ln|x - 1| + (3/2) ln (x^2+9) - (4/3) arctan (x/3) + c`

**Hence, evaluating the given integral yields `int (5x^2-7x+22)/((x-1)(x^2+9)) dx = 2 ln|x - 1| + (3/2) ln (x^2+9) - (4/3) arctan (x/3) + c.` **