# Integral of inverse trig or inverse hyperbolic

## Homework Statement

∫5/(4x√(9-16x2)dx

## Homework Equations

I am pretty sure this is in the form of ∫du/(u√(a2-u2)

## The Attempt at a Solution

setting u=4x a=3 and du=4dx so 1/4du=dx I get:

-5/12 sech-1(4x/3) + C

Is this right or am I using the wrong definition? Just trying to check my answers

Thanks for any help

HallsofIvy