Integrating $\frac{x}{y}$ & $(xy)^3$ - Get Help Now!

[paulmdrdo1]

how would you integrate

1. $\displaystyle \int \left(\frac{x}{y}\right) d\left(\frac{x}{y}\right)$

2. $\displaystyle \int \frac{d(xy)}{(xy)^3}$

if the two is just in this form I could easily answer it

$\displaystyle \int d\left(\frac{x}{y}\right)=(\frac{x}{y})$

$\displaystyle \int d(xy)=(xy)$

but I don't know how to treat $\frac{x}{y}$ in 1 and the $(xy)^3$ in 2.

please help!

 

[mente oscura]

how would you integrate

1. $\displaystyle \int \left(\frac{x}{y}\right) d\left(\frac{x}{y}\right)$

2. $\displaystyle \int \frac{d(xy)}{(xy)^3}$

if the two is just in this form I could easily answer it

$\displaystyle \int d\left(\frac{x}{y}\right)=(\frac{x}{y})$

$\displaystyle \int d(xy)=(xy)$

but I don't know how to treat $\frac{x}{y}$ in 1 and the $(xy)^3$ in 2.

please help!

Hello.

$$1) \ u=\dfrac{x}{y}$$

$$\displaystyle \int u \ d(u)=u= \dfrac{x}{y}$$

$$2) \ u=(xy)$$

$$\displaystyle \int \dfrac{d(u)}{u^3}=?$$

Can you follow?

Regards.

 

[MarkFL]

Hello.

$$1) \ u=\dfrac{x}{y}$$

$$\displaystyle \int u \ d(u)=u= \dfrac{x}{y}$$

What we want here is:

\(\displaystyle \int u\,du=\frac{1}{2}u^2+C\)

 

[mente oscura]

What we want here is:

\(\displaystyle \int u\,du=\frac{1}{2}u^2+C\)

Right. I I'm wrong.

I wrote:

$$\int u\,du$$

And I thought in:

$$\int \,du$$

(Headbang)(Headbang)(Headbang)

I'm sorry.

Regards.

 

[MarkFL]

Hey, we all make mistakes...I know I have made my share! (Emo)