Can anyone here explain how did he get it.

[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

 

[g_edgar]

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

 

[coki2000]

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

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

[tex]\int \frac{u^2}{u+3}\frac{du}{u}=\int \frac{u}{u+3}du[/tex]

 

[sarah22]

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

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

 

[coki2000]

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

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