# Can anyone here explain how did he get it.

1. Nov 30, 2009

### sarah22

I thought it is (e^x)^2 with u = e^x
so it should be u^2 but he get only u without a square. Can anyone here explain it.

The red box.
http://img6.imageshack.us/img6/3272/whatj.png [Broken]

Last edited by a moderator: May 4, 2017
2. Nov 30, 2009

### g_edgar

Put in between the red box and the previous line:
$$\int\frac{u}{u+3}\,du$$

3. Nov 30, 2009

### coki2000

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Hi
Because $$\int \frac{(e^x)^2}{e^x+3}dx\rightarrow u=e^x\Rightarrow du=e^xdx \Rightarrow dx=\frac{du}{e^x}$$
So,

$$\int \frac{u^2}{u+3}\frac{du}{u}=\int \frac{u}{u+3}du$$

Last edited by a moderator: May 4, 2017
4. Nov 30, 2009

### sarah22

LMAO. I forgot the dx part. woah. Laziness is here.

Anyway, Thanks. I got it now. (^_^)

5. Nov 30, 2009