1. ## Integral

Hi,

How to calculate the following integral $\int_{e^{3}}^{e^{4}}\frac{ln(1+\sqrt{x})}{x+\sqrt{ x}}$ ???

$\int{u*dv}=uv-\int{v*du}$

$u=ln(1+\sqrt{x})$
$du=\frac{1}{2\sqrt{x}*(\sqrt{x}+1)}$
$v=2ln(1+\sqrt{x})$
$dv=\frac{1}{x+\sqrt{x}}$

then

$uv=(ln(1+\sqrt{x}))*(2ln(1+\sqrt{x}))=2ln(1+\sqrt{ x})^2$

and

$\int{v*du}=\int{\frac{ln(1+\sqrt{x})}{\sqrt{x}(1+\ sqrt{x}}}$

Do it again and you'll get

$\int{\frac{ln(1+\sqrt{x})}{\sqrt{x}(1+\sqrt{x}}}=( integration by parts)=ln(1+\sqrt{x})^2)$

and so

$(\int{u*dv}=uv-\int{v*du})=((2ln(1+\sqrt{x})^2)-ln(1+\sqrt{x})^2)))=ln(1+\sqrt{x})^2$

And of course, now that you have done the integral you can enter the parameters and evaluate it.

3. $\int \frac{\ln(1+\sqrt{x})}{x+\sqrt{x}}~dx$

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

$= 2 \int \frac{\ln(1+\sqrt{x})}{1+\sqrt{x}}~d(\sqrt{x})$

$= 2\int \ln(1+\sqrt{x}) ~d(\ln(1+\sqrt{x}))$

sub. $u = \ln(1+\sqrt{x})$

The integral

$= 2 \int udu = u^2 + C = \ln^2(1+\sqrt{x}) + C$