Integration Problem

[Eternal Sky]

I am attempting to integrate the problem

sqrt(t) / [1 + sqrt(t)] dt

I've tried several different substitutions, but I cannot seem to come up with anything that I can solve.

If someone could help me out, I would greatly appreciate it.

 

[soroban]

Hello, Eternal Sky!

\(\displaystyle \text{Integrate: }\;\int\frac{\sqrt{t}}{1 + \sqrt{t}}\,dt\)

\(\displaystyle \text{Let }\,\sqrt{t} \:=\:u \quad\Rightarrow\quad t \:=\:u^2 \quad\Rightarrow\quad dt \:=\:2u\,du\)

\(\displaystyle \text{Substitute: }\:\int\frac{u}{1+u}(2u\,du) \;=\;2\int\frac{u^2}{u+1}\,du \;=\;2\int\left(u - 1 + \frac{1}{u+1}\right)\,du\)

Can you finish it now?

 

[Eternal Sky]

I can get the answer now, yes. However, I'm not sure I understand this step:

\(\displaystyle \;2\int\frac{u^2}{u+1}\,du \;=\;2\int\left(u - 1 + \frac{1}{u+1}\right)\,du\)

Could you explain where the u - 1 comes from?

 

[pka]

Do you know how to divide \(\displaystyle u^2\) by \(\displaystyle u+1\)?

 

[Eternal Sky]

Ah, now I understand.

Thank you both so much for your help!