Evaluate ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{10x^2-9x16}{x^{3}4x}dx$

Solution

Since $x^{3}4x=x(x^{2}4)$ can't be factored further, we write

$\frac{10x^2-9x16}{x(x^{2}4)}=\frac{A}{}\frac{BxC}{}$

Multiplying by $x(x^{2}4),$ we have

Equating coefficients, we obtain

$AB=10,C=-9,$ and $4A=16$

Therefore $A=,B=$ and $C=-9,$ and so

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{10x^2-9x16}{x^{3}4x}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(\frac{?}{x}\frac{()x-9}{x^{2}4})dx$

In order to integrate the second term we split it into two parts.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x-9}{x^{2}4}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{({)}^x}{x^{2}4}dx-{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{9}{x^{2}4}dx$

We make the substitution $u=x^{2}4$ in the first of these integrals so that $du=2xdx.$ We evaluate the second integral by means of