# How do you integrate # (x+1)/( (x^2 )*(x-1) )# using partial fractions?

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Continuation:

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To integrate ( \frac{x+1}{x^2(x-1)} ) using partial fractions, follow these steps:

- Express the fraction as a sum of simpler fractions using partial fraction decomposition.
- Determine the unknown constants.
- Integrate each term separately.

Given fraction: ( \frac{x+1}{x^2(x-1)} )

Step 1: Partial fraction decomposition:

[ \frac{x+1}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} ]

Step 2: Determine the unknown constants ( A, B, ) and ( C ):

Multiply both sides by the denominator to clear fractions:

[ x + 1 = A(x-1) + Bx(x-1) + Cx^2 ]

Expand and group like terms:

[ x + 1 = Ax - A + Bx^2 - Bx + Cx^2 ]

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

Compare coefficients of corresponding powers of ( x ):

For ( x^2 ): ( B + C = 0 )

For ( x ): ( A - B = 1 )

For the constant term: ( -A = 1 )

Solve this system of equations to find the values of ( A, B, ) and ( C ).

Step 3: Integrate each term separately once you've found ( A, B, ) and ( C ).

[ \int \frac{x+1}{x^2(x-1)} , dx = \int \left( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} \right) , dx ]

Integrate each term separately:

[ \int \frac{A}{x} , dx + \int \frac{B}{x^2} , dx + \int \frac{C}{x-1} , dx ]

[ = A\ln|x| - \frac{B}{x} + C\ln|x-1| + D ]

Where ( D ) is the constant of integration.

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