\int \frac{(x^4 - x^3 - 5x + 41)dx}{x^3 - 2x^3 + 3x - 6}


{eq}\int \frac{(x^4 - x^3 - 5x + 41)dx}{x^3 - 2x^3 + 3x - 6} {/eq}

Partial Fraction Decomposition

A proper rational function (a ratio of polynomials in which the polynomial in the numerator has degree less than the polynomial in the denominator) can be written as a sum of simpler functions using partial fraction decomposition. The denominator of the rational function should be fully factored into irreducible factors. If there is a linear factor {eq}x-a {/eq}, then there is a corresponding term in the partial fraction expansion {eq}\frac{A}{x-a} {/eq}. If a linear factor occurs n times, there are n corresponding terms in the partial fraction expansion, one for each power {eq}\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \ldots + \frac{A_n}{(x-a)^n} {/eq}. Similarly for any irreducible quadratic factor {eq}ax^2+bx+c {/eq} in the denominator, there is a corresponding term in the expansion {eq}\frac{Ax+B}{ax^2+bx+c} {/eq}. For quadratic factors that appear n times, there are n such factors - one for each power. Once the form of the partial fraction expansion is written out, all of the constants can be solved for by multiplying both sides of our equation by the denominator and then comparing coefficients on both sides of the equation. Partial fraction decomposition is used in calculus to rewrite integration problems so that they are easier to evaluate.

Answer and Explanation:

Note that the denominator in the way the problem was stated was {eq}x^3-2x^3+3x-6 {/eq} , which is a typo. It should be {eq}x^3-2x^2+3x-6 {/eq} as that denominator factors for this partial fraction problem

{eq}\int \frac{(x^4 - x^3 - 5x + 41)dx}{x^3 - 2x^2 + 3x - 6} {/eq}

Begin by long dividing the polynomials, since the numerator has higher degree then the denominator. After long dividing we have

{eq}\int \frac{(x^4 - x^3 - 5x + 41)dx}{x^3 - 2x^2 + 3x - 6} = \int (x+1 +\frac{-x^2+3x+47}{x^3-2x^2+3x-6}) dx {/eq}

For the last term, rewrite using the partial fraction decomposition. Begin by factoring the denominator and writing out the partial fraction decomposition with placeholders for the constants

{eq}\frac{-x^2+3x+47}{(x-2)(x^2+3)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+3}\\ -x^2+3x+47 = A(x^2+3) + (Bx+C)(x-2) {/eq}

Using x=2, we can easily find A

{eq}49=7A\\ 7=A {/eq}

Substituting A=7 back in, we can now distribute and compare coefficients to find B and C

{eq}-x^2+3x+47 = 7(x^2+3) + (Bx+C)(x-2)\\ -x^2+3x+47 = (7+B)x^2 + (-2B+C)x +21-2C {/eq}

So {eq}B=-8 , C=-13 {/eq}

Using the decomposition to rewrite the integral, we have

{eq}\int (x+1 +\frac{7}{x-2} -\frac{8x}{x^2+3}-\frac{13}{x^2+3}) dx = \frac{1}{2}x^2+x+7\ln|x-2|-4\ln|x^2+3|-\frac{13}{\sqrt{3}}\tan^{-1}(\frac{x}{\sqrt{3}}) +c {/eq}

Learn more about this topic:

Partial Fraction Decomposition: Rules & Examples

from High School Algebra I: Help and Review

Chapter 3 / Lesson 25

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