EXAMPLE $5$ Evaluate ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{7x^2-6x+16}{x^3+4x}dx$

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

$\frac{7x^2-6x+16}{x(x^2+4)}=+\frac{Bx+C}{?}$

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

$7x^2-6x+16=A(x^2+4)+()x$

$=$

Equating coefficients, we obtain

$A+B=7,C=-6,4A=16$

Thus $A=,B=,$ and $C=-6$ and so

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{7x^2-6x+16}{x^3+4x}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(\frac{4}{x}+\frac{3}{x^2+4})dx$

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

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{3,x-6}{x^2+4}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{3}{x^2+4}dx-{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{6}{x^2+4}dx$

We make the substitution $u=x^2+4$ in the first of these integrals so that $du=2xdx.$ We evaluate the seco means of this formula with $a=2.($Remember to use absolute values where appropriate.$)$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{7x^2-6x+16}{x(x^2+4)}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{4}{x}dx+{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(,)dx-{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{6}{x^2+4}dx$

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