Question

I am having trouble finding the indef integral sqrt(3x)/(1+sqrt(3x)). I understand u = sqrt(3x+10 and sqrt(3x) = u-1 and du = sqrt(3)/2sqrt(x). I am just not sure where to go from here.

Answer 1

I would approach \(\displaystyle \int \frac{\sqrt{3x}}{1+\sqrt{3x}}\,dx \) in the following way:
Let \(u=\sqrt{3x}+1\implies \sqrt{3x} = u-1\).
Therefore, \(\,du = \dfrac{3}{2\sqrt{3x}}\,dx \implies \,dx = \frac{2}{3}\sqrt{3x} \,du\).
Thus, \[\begin{aligned}\int\frac{\sqrt{3x}}{1+\sqrt{3x}}\,dx \xrightarrow{u=\sqrt{3x}+1}{} & \phantom{=}\frac{2}{3} \int\frac{(\sqrt{3x})^2}{u}\,du\\ & = \frac{2}{3}\int \frac{(u-1)^2}{u}\,du\\ &= \frac{2}{3}\int\frac{u^2-2u+1}{u}\,du \\ &= \frac{2}{3}\int u-2+\frac{1}{u}\,du\end{aligned}\] Can you take things from here? :-)

Answer 2

one thing i am not catching is in the numerator you go from \[\sqrt{3x} \to (\sqrt{3x})^2\] why is it squared

Answer 3

You want to note that making the substitution mentioned above yields
\[\begin{aligned}\int\frac{\sqrt{3x}}{1+\sqrt{3x}}\,dx &= \int\frac{\sqrt{3x}}{u}\cdot\frac{2}{3}\sqrt{3x}\,du\\ &= \frac{2}{3}\int \frac{(\sqrt{3x})^2}{u}\,du \\ &= \frac{2}{3}\int \frac{(u-1)^2}{u}\,du\\ \end{aligned}\] Does this clarify things? :-)

Answer 4

ahh ok. Thank you for your help.

Answer 5

You're welcome!