How do you integrate #int (x^3-x^2+1) / (x^4-x^3)# using partial fractions?

Answer 1

The answer is #=1/(2x^2)+1/x+ln(|x-1|)+C#

Divide the breakdown into partial fractions.

#(x^3-x^2+1)/(x^4-x^3)=(x^3-x^2+1)/(x^3(x-1))#
#=A/(x^3)+B/(x^2)+C/(x)+D/(x-1)#
#=(A(x-1)+B(x(x-1))+C(x^2(x-1))+D(x^3))/((x^4-x^3))#

Compare the numerators; the denominators are the same.

#(x^3-x^2+1)=A(x-1)+B(x(x-1))+C(x^2(x-1))+D(x^3)#
Let #x=0#, #=>#, #1=-A#, #=>#, #A=-1#
Coefficients of #x#
#0=A-B#, #=>#, #B=A=-1#
Coefficients of #x^2#
#-1=B-C#, #=>#, #C=B+1=-1+1=0#
Coefficients of #x^3#
#1=C+D#, #D=1-C=1-0=1#

Consequently,

#(x^3-x^2+1)/(x^4-x^3)=-1/(x^3)-1/(x^2)+0/(x)+1/(x-1)#

So,

#int((x^3-x^2+1)dx)x^4-x^3)=-int(1dx)/(x^3)-int(1dx)/(x^2)+int(1dx)/(x-1)#
#=1/(2x^2)+1/x+ln(|x-1|)+C#
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Answer 2

To integrate the given rational function using partial fractions, you first decompose it into simpler fractions. The partial fraction decomposition for (\frac{{x^3 - x^2 + 1}}{{x^4 - x^3}}) is:

(\frac{{x^3 - x^2 + 1}}{{x^4 - x^3}} = \frac{{A}}{{x}} + \frac{{B}}{{x^2}} + \frac{{C}}{{x - 1}} + \frac{{D}}{{x^2 - 1}})

After finding the values of A, B, C, and D, you can integrate each term separately. The solution involves finding the values of A, B, C, and D by equating coefficients and then integrating each term individually.

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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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