Question

Show all steps for setting up the partial fractions for
You DO NOT have to solve the integral. Please show all work to receive points. Thank you for your help.

Answer 1

Suppose we have a fraction of polynomials to be integrated as below: 

$P(x)=(2x^3+9)(x^2-3x+6)(x+2)\\\\Q(x)=(x-5)^2(x^2+4)^2\\\\ I= \int\limits^{}_{} {\frac{P(x)}{Q(x)}} \, dx$

First set up the partial fractions as follows.

      Since the highest degree term in the polynomial in P(x) is 2 x⁶ and the highest degree term in Q(x) is x⁵,  quotient is 2 x.  Now split the fraction into simpler fractions with each term containing only one factor of Q(x) or its powers. Then in each term, select the expression in the numerator as a coefficient multiplied by the derivative of the denominator.

$\frac{P(x)}{Q(x)}=\frac{(2x^3+9)(x^2-3x+6)(x+2)}{(x-5)^2(x^2+4)^2}\\\\=2x+\frac{A}{x-5}+\frac{B}{(x-5)^2}+\frac{Cx}{(x^2+4)}+\frac{Dx}{(x^2+4)^2}\\\\=$

$P(x)=[2x(x-5)+A](x^2+4)^2(x-5)+B(x^2+4)^2\\.\ \ \ \ \ +[ Cx(x^2+4) + Dx ] (x-5)^2\\\\Compare\ similar\ terms\ and\ write\ simultaneous\ linear\ equations\ in\\.\ \ \ A,B,C\ and\ D.\ \ and\ Solve\ for\ A,B,C, and\ D.\\\\Now$

$I= \int\limits^{}_{} {\frac{P(x)}{Q(x)}} \, dx =\\\\ \int\limits^{}_{} {2x} \, dx +\int\limits^{}_{} {\frac{A}{x-5}} \, dx +\int\limits^{}_{} {\frac{B}{(x-5)^2}} \, dx +\int\limits^{}_{} {\frac{Cx}{(x^2+4)}} \, dx +\int\limits^{}_{} {\frac{Dx}{(x^2+4)^2}} \, dx$

$I=x^2+A\ Log|x-5|-\frac{B}{x-5}+\frac{C}{2}\ Log (x^2+4)-\frac{D}{2(x^2+4)}+K$

The main idea is the the partial fractions should be such that the numerator is in the format of the derivative of the factor in the denominator.  Before that, the denominator must be split in to smallest factors.