Proper substitution help for integration appreciated.

$$\int\frac{\sqrt{2x - 1}}{2x + 3}dx$$

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.

• Suggestion; Substitute $\sqrt{2x-1}=u$! – zxcvber Jun 13 '16 at 10:50
• Ths substitution suggested by @zxcvber gives a rational function, so you get rid of the square root (in contrary to what you say in the question). Try it, and if you have problems, update your question with details, and you will get help... – mickep Jun 13 '16 at 10:52
• Thanks yes I get it now :) – fella27 Jun 13 '16 at 10:59

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 ?

• Yes, thank you! – fella27 Jun 13 '16 at 11:42

let $2x-1=u^2$ we have $$I=\int{\frac{\sqrt{2x-1}}{2x+3}}dx=\int{\frac{{{u}^{2}}}{{{u}^{2}}+4}}du=\int{\left( 1-\frac{4}{{{u}^{2}}+4} \right)}du=u-2{{\tan }^{-1}}\left( \frac{u}{2} \right)$$ $$I=\sqrt{2x-1}-2{{\tan }^{-1}}\left( \frac{\sqrt{2x-1}}{2} \right)$$

If we were to substitute; let $\sqrt{2x-1}=u$

Differentiate both sides to get $$\frac{1}{\sqrt{2x-1}}dx=du$$

Since we let $\sqrt{2x-1}=u$, $dx=udu$. Also, squaring both sides gives us $2x-1=u^2$, so $2x+3=u^2+4$.

Thus, $$\int\frac{\sqrt{2x-1}}{2x+3}dx=\int\frac{u^2}{u^2+4}du$$

Now, calculation...

$$\int\frac{u^2}{u^2+4}du=\int\frac{u^2+4-4}{u^2+4}du=\int \left( 1-\frac{4}{u^2+4}\right)du$$

I hope this helped.