Thread: Integrals - ln, sqrts

My English-Math language isn't too good, so I'll try to use math to explain:

$\displaystyle \int \sqrt{x} \ln{(1+\sqrt{x})} dx$

$\displaystyle t:=ln(1+\sqrt{x}) \Rightarrow dt=\frac{1}{2\sqrt{x}(1+\sqrt{x})}
\Rightarrow dx=2\sqrt{x}(1+\sqrt{x})dt$
$\displaystyle \Downarrow$
$\displaystyle e^t=1+\sqrt{x} \Rightarrow x=(e^t-1)^2$

$\displaystyle \int \sqrt{x} \ln{(1+\sqrt{x})} dx$ = $\displaystyle \int \sqrt{x} t*2\sqrt{x}(1+\sqrt{x})dt$

= $\displaystyle \int 2t(x+x^\frac{3}{2}) dt$

=$\displaystyle \int 2t((e^t-1)^2+(e^t-1)^3) dt $

=$\displaystyle \int 2t*e^t(e^t-1)^2)) dt $

Now what ?

Thank you very much

Originally Posted by adam63

My English-Math language isn't too good, so I'll try to use math to explain:

$\displaystyle \int \sqrt{x} \ln{(1+\sqrt{x})} dx$

$\displaystyle t:=ln(1+\sqrt{x}) \Rightarrow dt=\frac{1}{2\sqrt{x}(1+\sqrt{x})}
\Rightarrow dx=2\sqrt{x}(1+\sqrt{x})dt$
$\displaystyle \Downarrow$
$\displaystyle e^t=1+\sqrt{x} \Rightarrow x=(e^t-1)^2$

$\displaystyle \int \sqrt{x} \ln{(1+\sqrt{x})} dx$ = $\displaystyle \int \sqrt{x} t*2\sqrt{x}(1+\sqrt{x})dt$

= $\displaystyle \int 2t(x+x^\frac{3}{2}) dt$

=$\displaystyle \int 2t((e^t-1)^2+(e^t-1)^3) dt $

=$\displaystyle \int 2t*e^t(e^t-1)^2)) dt $

Now what ?

Thank you very much

Click on Show steps: integrate Sqrt[x]Log[1 + Sqrt[x]] - Wolfram|Alpha

Let $\displaystyle u = \sqrt{x} + 1$ so your integral becomes

$\displaystyle
2 \int (u-1)^2 \ln u \, du
$.

Expand and integrate by parts.

Thanks for the help I got it!