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

This allows us to write an equation system.

As a result, the partial fraction breakdown is as follows:

Here is one way to integrate this:

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To integrate ( \frac{2}{{x^3 - x^2}} ) using partial fractions, first factor the denominator, then decompose the fraction into partial fractions. In this case:

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

We can rewrite this as:

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

Multiplying both sides by the denominator ( x^2(x - 1) ), we get:

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

Now, we can solve for A, B, and C by comparing coefficients:

( 2 = Ax - A + Bx^2 - Bx + Cx^2 )

Equating coefficients of like terms, we get:

( 0x^2: B + C = 0 ) ( 1x: A - B = 0 ) ( 0x: -A = 2 )

Solving these equations simultaneously, we find:

( A = -2 ) ( B = -2 ) ( C = 2 )

Therefore, the partial fraction decomposition is:

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

Now, we can integrate each partial fraction separately:

( \int \frac{-2}{x} , dx = -2 \ln|x| + C_1 )

( \int \frac{-2}{x^2} , dx = 2 \frac{1}{x} + C_2 )

( \int \frac{2}{x - 1} , dx = 2 \ln|x - 1| + C_3 )

Finally, combining the results, we get:

( \int \frac{2}{{x^3 - x^2}} , dx = -2 \ln|x| + 2 \frac{1}{x} + 2 \ln|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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