# Thread: Integrals - ln, sqrts

1. ## Integrals - ln, sqrts

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

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

$t:=ln(1+\sqrt{x}) \Rightarrow dt=\frac{1}{2\sqrt{x}(1+\sqrt{x})}
\Rightarrow dx=2\sqrt{x}(1+\sqrt{x})dt$

$\Downarrow$
$e^t=1+\sqrt{x} \Rightarrow x=(e^t-1)^2$

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

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

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

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

Now what ?

Thank you very much

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

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

$t:=ln(1+\sqrt{x}) \Rightarrow dt=\frac{1}{2\sqrt{x}(1+\sqrt{x})}
\Rightarrow dx=2\sqrt{x}(1+\sqrt{x})dt$

$\Downarrow$
$e^t=1+\sqrt{x} \Rightarrow x=(e^t-1)^2$

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

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

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

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

Now what ?

Thank you very much
Click on Show steps: integrate Sqrt&#x5b;x&#x5d;Log&#x5b;1 &#x2b; Sqrt&#x5b;x&#x5d;&#x5d; - Wolfram|Alpha

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

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

Expand and integrate by parts.

4. Thanks for the help I got it!