Integrals - ln, sqrts

**adam63** · April 20th 2010, 04:32 AM

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

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

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

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

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

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

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

Now what

?

Thank you very much

**mr fantastic** · April 20th 2010, 04:36 AM

Originally Posted by adam63

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

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

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

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

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

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

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

Now what

?

Thank you very much

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

**Danny** · April 20th 2010, 04:38 AM

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

$<br /> 2 \int (u-1)^2 \ln u \, du<br />$ .

Expand and integrate by parts.

**adam63** · April 20th 2010, 05:13 AM

Thanks for the help

I got it!

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