Hi all,
I need help solving for this:
$\displaystyle
\int \frac{du}{3u + \sqrt{u}} = \int \frac{dx}{x}
$
Thanks in advance
ArTiCk
Let $\displaystyle \xi=\sqrt{u}\Rightarrow{\xi^2=u}$
So $\displaystyle du=2\xi{d\xi}$
So we have
$\displaystyle \int\frac{2\xi}{3\xi^2+\xi}d\xi=\frac{2\ln(3\xi+1) }{3}$
back subbing we get
$\displaystyle \frac{2\ln(3\sqrt{u}+1)}{3}=\ln((3\sqrt{u}+1)^{\fr ac{2}{2}})$
The right side is obviously $\displaystyle \ln(x)$
so we have
$\displaystyle \ln((3\sqrt{u}+1)^{\frac{2}{3}})=\ln(x)$
exponentiating both sides gives
$\displaystyle (3\sqrt{u}+1)^{\frac{2}{3}}=x$
and I assume this was solving for x
so done
Hello, ArTiCK!
Let $\displaystyle \sqrt{u} \,=\,v\quad\Rightarrow\quad u \,=\,v^2\quad\Rightarrow\quad du \,=\,2v\,dv$$\displaystyle \int \frac{du}{3u + \sqrt{u}} \:= \:\int \frac{dx}{x}$
Substitute: .$\displaystyle \int\frac{2v\,dv}{3v^2+v} \;=\;\int\frac{dx}{x} \quad\Longrightarrow\quad 2\int\frac{dv}{3v+1} \:=\:\int\frac{dx}{x}$
Go for it!