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

Answer 1

#=2(ln|x - 1| - ln|x| + 1/x) + C#

#x^3 - x^2# can be factored as #x^2(x - 1)#.
#(Ax + B)/(x^2) + C/(x - 1) = 2/((x^2)(x - 1))#
#(Ax + B)(x- 1) + C(x^2) = 2#
#Ax^2 + Bx - Ax - B + Cx^2 = 2#
#(A + C)x^2 + (B - A)x + (-B) = 2#

This allows us to write an equation system.

#{(A + C = 0), (B - A = 0), (-B = 2):}#
Solving, we obtain #B = -2#, #A = -2#, #C = 2#.

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

#(-2x - 2)/x^2 + 2/(x - 1)#
We now decompose #(-2x - 2)/x^2# into partial fractions.
#A/x^2 + B/x = (-2x- 2)/x^2#
#A + Bx = -2x - 2 -> A = -2 and B = -2#
Therefore, the complete partial fraction decomposition of #2/(x^3 - x^2)# is #-2/x^2 - 2/x + 2/(x - 1)#.

Here is one way to integrate this:

#int(-2/x^2 - 2/x + 2/(x - 1))dx#
#=-2ln|x| + 2ln|x- 1| - int(2x^-2) + C#
#=2ln|x - 1| - 2ln|x| -(-2x^-1) + C#
#=2ln|x- 1| - 2ln|x| + 2/x + C#
#=2(ln|x - 1| - ln|x| + 1/x) + C#

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Answer 2

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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Answer from HIX Tutor

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.

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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