1. ## Integral

Hi,

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

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

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

then

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

and

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

Do it again and you'll get

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

and so

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

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

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

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

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

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

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

The integral

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