Science:Math Exam Resources/Courses/MATH105/April 2016/Question 02 (c)/Solution 1

Since the degree of the numerator is smaller than the denominator which is 3 here, we use partial fractions to evaluate the integral.

The denominator is $x^{3}-9x^{2}=x^{2}(x-9)$, so we split the denominator as the following,

${\frac {81}{x^{3}-9x^{2}}}={\frac {A}{x^{2}}}+{\frac {B}{x}}+{\frac {D}{x-9}}.$

We find the numerator by computing the common denominator

${\frac {81}{x^{3}-9x^{2}}}={\frac {A}{x^{2}}}+{\frac {B}{x}}+{\frac {D}{x-9}}={\frac {A(x-9)+Bx(x-9)+Dx^{2}}{x^{2}(x-9)}}.$

We simplify the numerator and make it equal to $81$ to get

$(B+D)x^{2}+(A-9B)x-9A=81.$

Matching the coefficient, we find the following equations for $A,B,C$:

$B+D=0,A-9B=0,-9A=81$ which gives:

$A=-9,B=-1,D=1$, hence

$\int {\frac {81}{x^{3}-9x^{2}}}dx=-\int {\frac {9}{x^{2}}}dx-\int {\frac {1}{x}}dx+\int {\frac {1}{x-9}}dx={\frac {9}{x}}-\ln |x|+\ln |x-9|+C.$