Science:Math Exam Resources/Courses/MATH101/April 2010/Question 03 (a)/Solution 2

We want to evaluate the integral

$\int {\frac {x^{2}-9}{x(x^{2}+9)}}\,dx.$

As a first step, we split up the integral:

$\int {\frac {x^{2}-9}{x(x^{2}+9)}}\,dx=\int {\frac {x^{2}}{x(x^{2}+9)}}\,dx-\int {\frac {9}{x(x^{2}+9)}}\,dx$

We can evaluate the first integral by canceling a factor of x, and then substitute u= x^2+9, du = 2dx:

${\begin{aligned}\int {\frac {x^{2}}{x(x^{2}+9)}}\,dx&=\int {\frac {x}{x^{2}+9}}\,dx=\int {\frac {1}{2u}}\,du\\&={\frac {1}{2}}\ln |u|+C={\frac {1}{2}}\ln(x^{2}+9)+C\end{aligned}}$

The second integral can be evaluated using a partial fraction decomposition

${\frac {-9}{x(x^{2}+9)}}={\frac {A}{x}}+{\frac {Bx}{x^{2}+9}}+{\frac {C}{x^{2}+9}},$

which we can solve for $-9=A(x^{2}+9)+Bx^{2}+Cx.$ When x = -1, we get -9 = 10A+B-C, when x=1, we get -9 = 10A+B+C. Subtracting the two equations gives C = 0. Finally, when x = 2, we get -9 = 13A + 4B. So 10A + B = 13 A+ 4B, which yields A = -B. Thus -9 = 10A-A = 9A, so that A = -1 and B = 1. Therefore

$\int {\frac {-9}{x(x^{2}+9)}}\,dx=\int {\frac {-1}{x}}\,dx+\int {\frac {x}{x^{2}+9}}\,dx=-\ln |x|+C+\int {\frac {x}{x^{2}+9}}\,dx.$

We evaluated the last integral earlier in this problem and found $\int {\frac {x}{x^{2}+9}}\,dx={\frac {1}{2}}\ln(x^{2}+9).$ So

$\int {\frac {-9}{x(x^{2}+9)}}\,dx={\frac {1}{2}}\ln(x^{2}+9)-\ln |x|+C,$

and therefore the final answer is

$\int {\frac {x^{2}-9}{x(x^{2}+9)}}\,dx=\ln(x^{2}+9)-\ln |x|+C.$