# Compute the indefinite integral `int (2x^2-2x+1)/((x-1)(x^2+2)) dx` using partial fraction

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The integral `int (2x^2-2x+1)/((x-1)(x^2+2))dx` has to be determined.

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

=> `(Ax^2 + Bx - Ax - B + Cx^2 + 2C)/((x^2+2)(x-1))`

=> A + C = 2, B - A = -2 and 2C - B = 1

Add A + C = 2 and B - A = -2

=> B + C = 0

Substitute B = -C in 2C - B = 1

=> 3C = 1

=> C = 1/3

B = -1/3

A = 5/3

The integral is `int (1/3)(5x - 1)/(x^2 + 2) + (1/3)*1/(x-1) dx`

=> `(1/3)*int (5x)/(x^2 + 2)dx - (1/3)*int 1/(x^2 + 2) dx+ (1/3)*ln(x - 1) `

=> `(1/3)*int (5x)/(x^2+2)dx - (1/(3*sqrt2))*tan^-1(x/sqrt 2) + (ln(x-1))/3`

Let x^2 + 2 = y

dy/dx = 2x

=> (1/2)*dy = x*dx

=> `(1/3)*int (5/3)/(y) dy - (tan^-1(x/sqrt2))/(3*sqrt2) + (ln(x-1))/3`

=> `(5*ln(x^2 + 2))/9 - (tan^-1(x/sqrt2))/(3*sqrt2) + (ln(x-1))/3`

**The required integral is **` (5*ln(x^2+2))/9 - (tan^-1(x/sqrt 2)).(3*sqrt2) + (ln(x-1))/3 + C`