Using the method of partial fractions show your users that ∫ dx/x^2(x-2) = -1/4In(x) + 1/2x + 1/4In (x-2) + constant
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You need to use partial fraction decomposition to write the given integral as a sum (difference) of a simpler integrals such that:
`1/(x^2(x-2)) = A/x + B/x^2 + C/(x-2)`
`1 = Ax(x-2) + B(x-2) + Cx^2`
`1 = Ax^2 - 2Ax + Bx - 2B + Cx^2`
`1 =x^2(A+C) + x(-2A + B) - 2B`
Equating the coefficients of like powers yields:
`A+C = 0 => C = -A`
`-2A + B = 0 `
`-2B = 1 => B = -1/2`
Substituting `-1/2` for B in `-2A + B = 0 ` yields:
`-2A- 1/2 = 0 => -2A = 1/2 => A = -1/4 => C = 1/4`
`1/(x^2(x-2)) = -1/(4x) - 1/(2x^2) + 1/(4(x-2))`
Integrating both sides yields:
`int 1/(x^2(x-2)) dx= int -1/(4x) dx- int 1/(2x^2) dx+ int 1/(4(x-2))dx`
`int 1/(x^2(x-2)) dx =-(1/4)ln|x| -(1/2)(x^(-2+1))/(-2+1) + int 1/(4t)dt`
`int 1/(x^2(x-2)) dx = -(1/4)ln|x| + 1/(2x) + (1/4)ln|t| + c`
Substituting back `x - 2` for `t` yields:
`int 1/(x^2(x-2)) dx = -(1/4)ln|x| + 1/(2x) + (1/4)ln|x - 2| + c`
Hence, evaluating the given integral, using partial fraction decomposition, yields `int 1/(x^2(x-2)) dx = -(1/4)ln|x| + 1/(2x) + (1/4)ln|x - 2| + c.`
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