How do you find #int (5x^2+3x2)/(x^3+2x^2) dx# using partial fractions?
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To find ∫(5x^2 + 3x  2) / (x^3 + 2x^2) dx using partial fractions, follow these steps:

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

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

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

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)

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

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)

Integrate each term separately: ∫(3/2) / x dx = (3/2) lnx ∫(3/4) / (x + 2) dx = (3/4) lnx + 2 ∫(23/4) * x / (x + 2) dx = (23/4) * (∫dx  ∫2 / (x + 2) dx) = (23/4) * (x  2 lnx + 2)

Combine the integrals: ∫(5x^2 + 3x  2) / (x^3 + 2x^2) dx = (3/2) lnx  (3/4) lnx + 2 + (23/4) * (x  2 lnx + 2) + C
Therefore, the integral of (5x^2 + 3x  2) / (x^3 + 2x^2) dx using partial fractions is (3/2) lnx  (3/4) lnx + 2 + (23/4) * (x  2 lnx + 2) + C, where C is the constant of integration.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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