# How do you integrate #int x/ (x^3-2x^2+x)# using partial fractions?

Make the integrand simpler:

Divide the denominator by two:

Then:

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To integrate the expression ( \frac{x}{x^3 - 2x^2 + x} ) using partial fractions, follow these steps:

- Factor the denominator: ( x^3 - 2x^2 + x = x(x^2 - 2x + 1) = x(x - 1)^2 ).
- Write the partial fraction decomposition: ( \frac{x}{x(x - 1)^2} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2} ).
- Clear the fractions by multiplying both sides by the denominator: ( x = A(x - 1)^2 + Bx(x - 1) + Cx ).
- Expand and combine like terms: ( x = A(x^2 - 2x + 1) + B(x^2 - x) + Cx ).
- Expand further: ( x = Ax^2 - 2Ax + A + Bx^2 - Bx + Cx ).
- Match coefficients of like terms: ( x^2 ) terms: ( A + B = 0 ), ( x ) terms: ( -2A - B + C = 1 ), constant terms: ( A = 0 ).
- Solve the system of equations: ( A = 0 ), ( B = -1 ), ( C = -1 ).
- Write the partial fraction decomposition: ( \frac{x}{x(x - 1)^2} = \frac{-1}{x} + \frac{-1}{x - 1} + \frac{-1}{(x - 1)^2} ).
- Integrate each term: ( \int{\frac{x}{x(x - 1)^2}}dx = \int{\frac{-1}{x}}dx + \int{\frac{-1}{x - 1}}dx + \int{\frac{-1}{(x - 1)^2}}dx ).
- Evaluate the integrals: ( -\ln{|x|} - \ln{|x - 1|} + \frac{1}{x - 1} + C ).

So, the integral of ( \frac{x}{x^3 - 2x^2 + x} ) using partial fractions is ( -\ln{|x|} - \ln{|x - 1|} + \frac{1}{x - 1} + C ).

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To integrate ( \int \frac{x}{x^3 - 2x^2 + x} ) using partial fractions, first factor the denominator:

[ x^3 - 2x^2 + x = x(x^2 - 2x + 1) = x(x - 1)^2 ]

Now, express the fraction as a sum of simpler fractions:

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

Multiply both sides by the denominator ( x(x - 1)^2 ):

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

Expand and collect like terms:

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

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

Compare coefficients of like terms:

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

For ( x ): ( -2A - B + C = 1 )

For constants: ( A = 0 )

Solving this system of equations, we find ( A = 0 ), ( B = -1 ), and ( C = 1 ).

Now, rewrite the original fraction with these values:

[ \frac{x}{x^3 - 2x^2 + x} = \frac{-1}{x - 1} + \frac{1}{(x - 1)^2} ]

Now, integrate each term separately:

[ \int \frac{x}{x^3 - 2x^2 + x} dx = -\ln|x - 1| - \frac{1}{x - 1} + C ]

Therefore, the integral of ( \frac{x}{x^3 - 2x^2 + x} ) using partial fractions is:

[ -\ln|x - 1| - \frac{1}{x - 1} + C ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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