How do you find #int (5x^2+3x-2)/(x^3+2x^2) dx# using partial fractions?

Answer 1

#2 ln x + 3 ln (x+2) +1/x+C#

Assume that you can write the fraction as #(Ax+B)/x^2+C/(x+2)#. That is, #5x^2+3x-2-=(Ax+B)(x+2)+Cx^2#
Then, equating coefficents (or however you prefer) you get #5=A+C#, #3=2A+B# and #-2=2B#. Hence #B=-1#, #A=2# and #C=3# So now you have #int(2x-1)/x^2+3/x+2 dx# =#int2/x-1/x^2+3/(x+2)dx# giving the answer above.
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Answer 2

To find ∫(5x^2 + 3x - 2) / (x^3 + 2x^2) dx using partial fractions, follow these steps:

  1. Factor the denominator: x^3 + 2x^2 = x^2(x + 2)

  2. Decompose the fraction into partial fractions: (5x^2 + 3x - 2) / (x^3 + 2x^2) = A/x + B/(x + 2) + Cx/(x + 2)

  3. Multiply both sides by the denominator to clear the fraction: 5x^2 + 3x - 2 = A(x + 2) + Bx(x + 2) + Cx^2

  4. Expand and equate coefficients: 5x^2 + 3x - 2 = Ax + 2A + Bx^2 + 2Bx + Cx^2

    Coefficients of like terms must match: A + 2B = 0 (coefficients of x terms) B + C = 5 (coefficients of x^2 terms) 2A = 3 (constant terms)

  5. Solve the system of equations: From A + 2B = 0, we get A = -2B From 2A = 3, we get A = 3/2 Substitute A = 3/2 into A = -2B to find B: 3/2 = -2B → B = -3/4 Substitute B = -3/4 into B + C = 5 to find C: -3/4 + C = 5 → C = 23/4

  6. Write the partial fraction decomposition: (5x^2 + 3x - 2) / (x^3 + 2x^2) = (3/2) / x - (3/4) / (x + 2) + (23/4) * x / (x + 2)

  7. Integrate each term separately: ∫(3/2) / x dx = (3/2) ln|x| ∫(3/4) / (x + 2) dx = (3/4) ln|x + 2| ∫(23/4) * x / (x + 2) dx = (23/4) * (∫dx - ∫2 / (x + 2) dx) = (23/4) * (x - 2 ln|x + 2|)

  8. Combine the integrals: ∫(5x^2 + 3x - 2) / (x^3 + 2x^2) dx = (3/2) ln|x| - (3/4) ln|x + 2| + (23/4) * (x - 2 ln|x + 2|) + C

Therefore, the integral of (5x^2 + 3x - 2) / (x^3 + 2x^2) dx using partial fractions is (3/2) ln|x| - (3/4) ln|x + 2| + (23/4) * (x - 2 ln|x + 2|) + C, where C is the constant of integration.

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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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