Integrate Functions Where the Denominator Contains Irreducible Quadratic Factors - dummies

Integrate Functions Where the Denominator Contains Irreducible Quadratic Factors

By Mark Ryan

Sometimes you can’t factor a denominator all the way down to linear factors because some quadratics are irreducible — like prime numbers, they can’t be factored.

Check the discriminant. You can easily check whether a quadratic (ax2 + bx + c) is reducible or not by checking its discriminant, b2 – 4ac. If the discriminant is negative, the quadratic is irreducible. If the discriminant is a perfect square like 0, 1, 4, 9, 16, 25, etc., the quadratic can be factored into factors like you’re used to seeing like (2x – 5)(x + 5). The last possibility is that the discriminant equals a non-square positive number, as with the quadratic x2 + 10x + 1, for example, that has a discriminant of 96. In that case, the quadratic can be factored, but you get ugly factors involving square roots. Fortunately, these problems are rare.

Using the partial fractions technique with irreducible quadratics is a bit different. Here’s a problem: Integrate

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  1. Factor the denominator.

    It’s already done! Note that x2 + 4 is irreducible because its discriminant is negative.

  2. Break up the fraction into a sum of “partial fractions.”

    If you have an irreducible quadratic factor (like the x2 + 4), the numerator for that partial fraction needs two capital-letter unknowns instead of just one. You write them in the form of Px + Q.

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  3. Multiply both sides of this equation by the left-side denominator.

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  4. Take the roots of the linear factors and plug them — one at a time — into x in the equation from Step 3, and then solve.

    If x = 0,If x = 1,

    –4 = –4A10 = 5B

    A = 1B = 2

    You can’t solve for all the unknowns by plugging in the roots of the linear factors, so you have more work to do.

  5. Plug into the Step 3 equation the known values of A and B and any two values for x not used in Step 4 (low numbers make the arithmetic easier) to get a system of two equations in C and D.

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  6. Solve the system: 1 = –C + D and 7 = 2C + D.

    You should get C = 2 and D = 3.

  7. Split up the original integral and integrate.

    Using the values obtained in Steps 4 and 6, A = 1, B = 2, C = 2, and D = 3, and the equation from Step 2, you can split up the original integral into three pieces:

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    And with simple algebra, you can split up the third integral on the right into two pieces, resulting in the final partial fraction decomposition:

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    The first two integrals are easy. For the third, you use substitution with

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    The fourth is done with the arctangent rule, which you should memorize: