$$\int\frac{\sqrt{2x - 1}}{2x + 3}dx$$
I'm completely stumped. I feel like my goal should be to get rid of the square root somehow, but I'm not sure how to accomplish that. As it stands, obviously a substitution for $\sqrt{2x - 1}$ just moves it down to the denominator and does nothing close to help, substitutions for $2x - 1$ or $2x + 3$ don't work either. I've tried integration by parts taking the numerator and denominator to be $u$ and $dv$ respectively and vice versa but that seems to make it more complicated.
A trignometric substitution seems to be my best bet, for example I tried $x = \frac{sec^2(u)}{2}$ which does indeed get rid of the square root but then leaves me with $$\int\frac{du}{sec^4(u) + 3sec^2(u)}$$ Which I'm also not sure how to do... Nudges in the right direction appreciated greatly.
Answer
Your first idea was good : if you substitute $u=\sqrt{2x-1}$, you have :
$$\int\frac{\sqrt{2x - 1}}{2x + 3}dx=\int\frac{\sqrt{2x - 1}\sqrt{2x - 1}}{(2x + 3)\sqrt{2x - 1}}dx=\int\frac{u^2}{u^2+4}du$$
Because $du=\frac{dx}{\sqrt{2x-1}}$ and $2x+3=2x-1+4=u^2+4$.
Can you go from there ?
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