1. ## Integral

How do u integrate:

$\displaystyle\int \dfrac{\sqrt{x-3}}{x^2}\,dx$

Thanks

2. Originally Posted by polymerase
How do u integrate:

$\displaystyle\int \dfrac{\sqrt{x-3}}{x^2}\,dx$

Thanks
There might be an easier, more elegant, way to do this, but start with
$x = 3 tan^2(\theta)$
and see where that takes you.

-Dan

3. Originally Posted by polymerase
How do u integrate:

$\displaystyle\int \dfrac{\sqrt{x-3}}{x^2}\,dx$
Mmm. Integration by parts:

$u=\sqrt{x-3}\implies du=\frac1{2\sqrt{x-3}}\,dx$ & $dv=x^{-2}\,dx\implies v=-\frac1x.$ The integral becomes to

$\int {\frac{{\sqrt {x - 3} }}
{{x^2 }}\,dx} = - \frac{{\sqrt {x - 3} }}
{x} + \frac{1}
{2}\underbrace {\int {\frac{1}
{{x\sqrt {x - 3} }}\,dx} }_\varphi .$

$\varphi$ is an arctangent, just set $\alpha=\sqrt{x-3}$ to descover it.

4. Originally Posted by Krizalid
Mmm. Integration by parts:

$u=\sqrt{x-3}\implies du=\frac1{2\sqrt{x-3}}\,dx$ & $dv=x^{-2}\,dx\implies v=-\frac1x.$ The integral becomes to

$\int {\frac{{\sqrt {x - 3} }}
{{x^2 }}\,dx} = - \frac{{\sqrt {x - 3} }}
{x} + \frac{1}
{2}\underbrace {\int {\frac{1}
{{x\sqrt {x - 3} }}\,dx} }_\varphi .$

$\varphi$ is an arctangent, just set $\alpha=\sqrt{x-3}$ to descover it.
huh? I get the byparts but the $\alpha$ stuff what?

5. You have: $\int{\frac{dx}{x\cdot{2\cdot{\sqrt[]{x-3}}}}}$

Now let: $u=\sqrt[]{x-3}$ then $\frac{du}{dx}=\frac{1}{2\cdot{\sqrt[]{x-3}}}$

Thus we have: $\int{\frac{dx}{x\cdot{2\cdot{\sqrt[]{x-3}}}}}=\int{\frac{\left(\sqrt[]{x-3}\right)'}{x}\cdot{dx}}=\int{\frac{du}{u^2+3}}$

Which is clearly of the arctan form

$\int{\frac{du}{u^2+3}}=\frac{1}{3}\cdot{\int{\frac {du}{\left(\frac{u}{\sqrt[]{3}}\right)^2+1}}}=\frac{\sqrt[]{3}}{3}\cdot{\arctan (\frac{u}{\sqrt[]{3}}})+C$

Which means that $\int{\frac{dx}{x\cdot{2\cdot{\sqrt[]{x-3}}}}}=\frac{\sqrt[]{3}}{3}\cdot{\arctan (\frac{\sqrt[]{x-3}}{\sqrt[]{3}}})+C$