Here's the question you clicked on:
stottrupbailey
integrate 6x/(2x+1)
\[\int\limits_{}^{}\frac{ 6x }{ 2x+1 }dx\]
Try to rewrite the fraction: in the denominator, there is 2x+1. If you had that in the numerator as well, things would become easier. So, if it was: (6x+3)/(2x+1), this would be 3(2x+1)/(2x+1)=3. We cannot make it that simple, but we can go in the same direction:\[\int\limits_{}^{}\frac{ 6x }{ 2x+1 }dx=\int\limits_{}^{}\frac{ 6x+3-3 }{ 2x+1 }dx=\int\limits_{}^{}\frac{ 3(2x+1)-3 }{ 2x+1 }dx\]Now we can split it up:\[=\int\limits_{}^{}3dx-\int\limits_{}^{}\frac{ 3 }{ 2x+1 }dx\]Maybe you can try this yourself...
ah, I see... never would've gotten that lol thanks so much! :)
BTW, although you said you'd never think of this yourself, next time you will!