Question

integral of (1)/(xsqrt(4x^2-9))

Answer 1

... this is too hard 4 m3 what ish dis? so hard!!

Answer 2

use substituttion

let

\[u = 4x^2 - 9\]

and then

\[\frac{du}{dx} = 8x \]

so

\[\frac{1}{8} du = x ~ dx\]

so the substitution becomes

\[\int\limits \frac{1}{8} u^{-\frac{1}{2}} du~~or~~\frac{1}{8} \int\limits u^{-\frac{1}{2}}
~du\]

hope it helps

Answer 3

O.o

Answer 4

I'm actually suppose to use trig substitution on this problem. That's why I was confused