MHB 242.8.7.64 int (x^4+1)/(x^3+9x) dx

[karush]

$\text{206.8.7.64}$
$\text{Given and evaluation}$
$$\displaystyle
I_{64}=\int \frac{x^4+1}{x^3+9x} \, dx
=\dfrac{\ln\left(\left|x\right|\right)}{9}-\dfrac{41\ln\left(x^2+9\right)}{9}+\dfrac{x^2}{2}$$
$\text{expand (via TI)}$
$$I_{64}= \frac{1}{9}\int\frac{1 }{x} \, dx
-\frac{82}{9}\int\frac{x}{(x^2+9)}\, dx
+\int x \, dx $$
$\text{OK I can see how the integral was evaluted }$
$\text{just don't see how the expansion was done?}$

 

[MarkFL]

I would write:

$$\frac{x^4+1}{x^3+9x}=\frac{x^4+9x^2+1-9x^2}{x^3+9x}=\frac{x(x^3+9x)+1-9x^2}{x^3+9x}=x-\frac{9x^2-1}{x^3+9x}$$

Now, using partial fractions:

$$\frac{9x^2-1}{x^3+9x}=\frac{A}{x}+\frac{Bx+C}{x^2+9}$$

$$9x^2-1=A(x^2+9)+(Bx+C)x=(A+B)x^2+Cx+9A$$

Equating coefficients, we obtain:

$$A+B=9$$

$$C=0$$

$$9A=-1\implies A=-\frac{1}{9}\implies B=\frac{82}{9}$$

Hence:

$$\frac{9x^2-1}{x^3+9x}=-\frac{1}{9x}+\frac{82x}{9(x^2+9)}$$

Putting it all together:

$$\frac{x^4+1}{x^3+9x}=x+\frac{1}{9x}-\frac{82x}{9(x^2+9)}$$