Hello Archduke01 Originally Posted by

**Archduke01** $\displaystyle \int \frac{\sqrt {4x^2-16}}{x} dx

$

$\displaystyle = \int \frac {\sqrt{(2x)^2 - 4^2}} {x} dx$

2x = 4 sec t

x = 2 sec t

$\displaystyle \int \frac {\sqrt {4(2sect)^2 - 16}} {2sect} dx$

$\displaystyle \int \frac {\sqrt {16(sec^2t-1}}{2sect} dx$

$\displaystyle \int \frac {4 \sqrt {sec^2t-1}}{2sect} dx$

How can I proceed? I don't think there's an identity to simplify sec^2 t - 1.

You don't need the trig substitution, if you put$\displaystyle 4x^2 - 16 = u^2$

$\displaystyle \Rightarrow x\,dx = \tfrac14u\,du$

Then:$\displaystyle \int \frac{\sqrt {4x^2-16}}{x} dx$$\displaystyle =\int \frac{\sqrt {4x^2-16}}{x^2}\,x\,dx$

$\displaystyle =\int\frac{u}{\tfrac14(u^2+16)}\;\tfrac14u\,du$

$\displaystyle =\int\frac{u^2}{u^2+16}\;du$

$\displaystyle =\int\left(1-\frac{16}{u^2+16}\right)\;du$

Can you complete it now?

Grandad