Question

Evaluate the indefinite integral of:
1
----
1 + (t)^1/2

dt

I would like a hint. I know substitution is involved.

Answer 1

\[\frac{1}{1+\sqrt{t}}=\frac{1}{\sqrt{t}}-\frac{1}{\sqrt{t}(\sqrt{t}+1)}\]

Answer 2

Thank you.

Answer 3

I was able to get u = (t)^1/2, but then I could not find an expression for 1/2(t)^1/2

Answer 4

you could have also let \[u=\sqrt{t}\] and worked it out from there

Answer 5

\[du=\frac{1}{2\sqrt{t}}dt\]
\[2\sqrt{t}du=dt\]
\[2udu=dt\]

\[\int\frac{1}{1+\sqrt{t}}dt=\int\frac{2u}{1+u}du\]

Answer 6

\[=2\int\frac{u}{1+u}du=2\int\frac{1+u-1}{1+u}du\]

\[=2\int\left[\frac{1+u}{1+u}+\frac{-1}{1+u}\right]du=2\int\left[1+\frac{-1}{1+u}\right]du\]

\[=2(u-\ln(|1+u|)+c=2\sqrt{t}-2\ln(\sqrt{t}+1)+c
\]