- #1
karush
Gold Member
MHB
- 3,269
- 5
$\text{206.8.7.64}$
$\text{Given and evaluation}$
$$\displaystyle
I_{64}=\int \frac{x^4+1}{x^3+9x} \, dx
=\dfrac{\ln\left(\left|x\right|\right)}{9}-\dfrac{41\ln\left(x^2+9\right)}{9}+\dfrac{x^2}{2}$$
$\text{expand (via TI)}$
$$I_{64}= \frac{1}{9}\int\frac{1 }{x} \, dx
-\frac{82}{9}\int\frac{x}{(x^2+9)}\, dx
+\int x \, dx $$
$\text{OK I can see how the integral was evaluted }$
$\text{just don't see how the expansion was done?}$
![Hot beverage :coffee: ☕](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/2615.png)
$\text{Given and evaluation}$
$$\displaystyle
I_{64}=\int \frac{x^4+1}{x^3+9x} \, dx
=\dfrac{\ln\left(\left|x\right|\right)}{9}-\dfrac{41\ln\left(x^2+9\right)}{9}+\dfrac{x^2}{2}$$
$\text{expand (via TI)}$
$$I_{64}= \frac{1}{9}\int\frac{1 }{x} \, dx
-\frac{82}{9}\int\frac{x}{(x^2+9)}\, dx
+\int x \, dx $$
$\text{OK I can see how the integral was evaluted }$
$\text{just don't see how the expansion was done?}$
![Hot beverage :coffee: ☕](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/2615.png)