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

Answer 1

#int((y^2+1)dy)/(y^3-1)=2/3ln(y-1)+ln(y^2+y+1)/6-1/sqrt3arctan((2y+1)/sqrt3)#

Let's rewrite the denominator as #y^3-1=(y-1)(y^2+y+1)#
And the decompsition into partial fractions #(y^2+1)/(y^3-1)=A/(y-1)+(Bx+C)/(y^2+y+1)#
#=(A(y^2+y+1)+(By+C)(y-1))/((y-1)(y^2+y+1))#
So, #y^2+1=A(y^2+y+1)+(By+C)(y-1)# Let #y=0# then, #1=A-C# Coefficients of #y^2#, #1=A+B# Coefficients of #y#, #0=A-B+C# Solving for A, B and C, we get #A=2/3#, #B=1/3# and #C=-1/3# Then #int((y^2+1)dy)/(y^3-1)=int(2dy)/(3(y-1))+int((1/3y-1/3)dy)/(y^2+y+1#
#=int(2dy)/(3(y-1))+int((2y+1)dy)/(6(y^2+y+1))-int(dy)/(2(y^2+y+1))# We have to find 3 integrals #int(2dy)/(3(y-1))=2/3ln(y-1)# #int((2y+1)dy)/(6(y^2+y+1))=ln(y^2+y+1)/6#
And the last integral #int(dy)/(2(y^2+y+1))=1/2intdy/(y^2+y+1/4+3/4)# #=1/2intdy/((y+1/2)^2+3/4)# #=1/2*4/3intdy/(((y+1/2)/(sqrt3/2))^2+1)# Let #u=(y+1/2)/(sqrt3/2)=(2y+1)/sqrt3# then #du=2dy/sqrt3# #=>##dy=(sqrt3du)/2# so #=2/3int(sqrt3/2du)/(u^2+1)=1/sqrt3int(du)/(u^2+1)# #=1/sqrt3arctanu=1/sqrt3arctan((2y+1)/sqrt3)# So putting it all together #int((y^2+1)dy)/(y^3-1)=2/3ln(y-1)+ln(y^2+y+1)/6-1/sqrt3arctan((2y+1)/sqrt3)#
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Answer 2

To integrate (\frac{y^2 + 1}{y^3 - 1}) using partial fractions, follow these steps:

  1. Factor the denominator (y^3 - 1): [ y^3 - 1 = (y - 1)(y^2 + y + 1) ]

  2. Decompose the fraction into partial fractions: [ \frac{y^2 + 1}{y^3 - 1} = \frac{A}{y - 1} + \frac{By + C}{y^2 + y + 1} ]

  3. Clear the fractions by multiplying through by the common denominator (y^3 - 1): [ y^2 + 1 = A(y^2 + y + 1) + (By + C)(y - 1) ]

  4. Expand and collect like terms: [ y^2 + 1 = Ay^2 + Ay + A + By^2 - By + Cy - C ]

  5. Group the terms with like powers of (y): [ y^2 \text{ terms}: \quad A + B = 1 ] [ y \text{ terms}: \quad A - B + C = 0 ] [ \text{constant terms}: \quad A + C = 1 ]

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

From the first equation: [ A + B = 1 ] Given that (A = 1 - B).

From the third equation: [ A + C = 1 ] Given that (C = 1 - A).

Substituting (A = 1 - B) into (A + C = 1): [ (1 - B) + C = 1 ] [ C = B ]

Using this result in the second equation: [ (1 - B) - B + B = 0 ] [ B = 1 ]

Then, (A = 1 - B = 0) and (C = B = 1).

  1. Rewrite the original integral using the partial fractions: [ \int \frac{y^2 + 1}{y^3 - 1} , dy = \int \frac{0}{y - 1} , dy + \int \frac{y + 1}{y^2 + y + 1} , dy ]

  2. Integrate each term: [ \int \frac{y + 1}{y^2 + y + 1} , dy ] can be solved by completing the square or using a trigonometric substitution.

After integrating each term, you'll get the final result for the integral.

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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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