# Math Help - Integration Question

1. ## Integration Question

The Integral of $1/
(2x+3)$

I am wondering if I could pull out the 2 from the denominator? So, I'd end up with $1/(2(x+3/2))$ Then could I just do 1/2 the integral of $1/(x+(3/2))$ which is ln(x + 3/2)? so the answer would be 1/2 LN(x+ 3/2)?

Is there a better way, or a correct way to solve this?

Thank you

2. Just put $u=2x+3$ and you'll arrive at $\frac12\int\frac{du}u=\frac12\ln|u|+k.$ Just back-substitute.