# Integrate the following: A.) \int \frac{1}{x^2 +5x +6} dx B.) \int \frac{x^3 +6x^2 +3x + 6}{x^3...

## Question:

Integrate the following:

A.) {eq}\displaystyle \int \frac{1}{x^2 +5x +6} dx {/eq}

B.) {eq}\displaystyle \int \frac{x^3 +6x^2 +3x + 6}{x^3 - 2x^2} dx {/eq}

C.) {eq}\displaystyle \int \frac{x^3 +5x^2 +2x - 1}{x^4 - 1} dx {/eq}

## Integration Techniques - Partial Fractions

When integrating a rational function, it is often convenient to rewrite the integrand using partial fraction decomposition. Partial fraction decomposition is essentially the opposite of finding a common denominator to write as one fraction - instead, you split a fraction up into simpler terms. If you have a proper rational function, begin by factoring the denominator completely. If there is a linear factor in the denominator {eq}ax+b {/eq}, there is a corresponding term in the partial fraction expansion {eq}\frac{A}{ax+b} {/eq}. If this factor occurs multiple times, there is a term for each power of the factor. If there is an irreducible quadratic factor {eq}ax^2+bx+c {/eq}, then there is a corresponding term in the expansion {eq}\frac{Ax+B}{ax^2+bx+c} {/eq}, and if the factor appears multiple times, there is a term for each power of the factor. The constants can then be solved for using algebra. When you rewrite the integrand in its partial fraction decomposition, you have a sum of simpler functions to integrate.

A) {eq}\int \frac{1}{x^2+5x+6} dx = \int \frac{1}{(x+3)(x+2)} dx {/eq}

Find the partial fraction expansion of the integrand

{eq}\frac{1}{(x+3)(x+2)} = \frac{A}{x+3}+\frac{B}{x+2}\\ 1=A(x+2)+B(x+3) {/eq}
x = -2 yields B=1 and x= -3 yields A=-1. Rewriting the integral, we have

{eq}\int\frac{1}{(x+3)(x+2)}dx=\int\left(\frac{-1}{x+3}+\frac{1}{x+2}\right) dx = -\ln|x+3| +\ln|x+2| + c {/eq}

B) {eq}\int \frac{x^3 +6x^2 +3x + 6}{x^3 - 2x^2} dx {/eq}

Begin with polynomial long division. Rewriting the integral, we have

{eq}\int \frac{x^3 +6x^2 +3x + 6}{x^3 - 2x^2} dx = \int \left( 1+\frac{8x^2+3x+6}{x^3-2x^2}\right) dx {/eq}

Continue by finding the partial fraction decomposition of the second term

{eq}\frac{8x^2+3x+6}{x^2(x-2)} = \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-2}\\ 8x^2+3x+6 = Ax(x-2) + B(x-2) + Cx^2\\ {/eq} x=2 yields C=11 and x=0 yields B=-3. Substituting these values in, distributing and comparing coefficients gives

{eq}8x^2+3x+6 = (A+11)x^2 + (-2A-3)x +6\\ {/eq} , so A = -3. Rewriting the integral, we have

{eq}\int \left( 1+\frac{8x^2+3x+6}{x^3-2x^2}\right) dx = \int\left(1-\frac{3}{x} -\frac{3}{x^2}+\frac{11}{x-2}\right) dx = x-3\ln|x|+\frac{3}{x}+11\ln|x-2| +c {/eq}

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

Find the partial fraction expansion

{eq}\frac{x^3 +5x^2 +2x - 1}{(x^2+1)(x-1)(x+1)} = \frac{Ax+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{x+1}\\ x^3 +5x^2 +2x - 1=(Ax+B)(x-1)(x+1)+C(x+1)(x^2+1)+D(x-1)(x^2+1)\\ {/eq} x=1 yields {eq}C=\frac{7}{4} {/eq} and x= -1 yields {eq}D=-\frac{1}{4} {/eq}. Substituting these values in, distributing and comparing coefficients gives

{eq}x^3+5x^2+2x-1 = (A+\frac{3}{2})x^3+(B+2)x^2+(\frac{3}{2}-A)x+(2-B) {/eq}, so {eq}A=-\frac{1}{2} , B=3 {/eq}

Rewriting the integral,

{eq}\int \frac{x^3 +5x^2 +2x - 1}{(x^2+1)(x-1)(x+1)} dx = \int\left(-\frac{x}{2(x^2+1)}+\frac{3}{x^2+1}+\frac{7}{4(x-1)}-\frac{1}{4(x+1)}\right) dx = -\frac{1}{4}\ln|x^2+1|+3\tan^{-1}(x)+\frac{7}{4}\ln|x-1|-\frac{1}{4}\ln|x+1| +c {/eq}