Integral of inverse trig or inverse hyperbolic

[USN2ENG]

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]

Yes, that is correct.

 

[USN2ENG]

Thanks, I really appreciate it!

One quick follow-up question to anyone who can help:

Some of the integral definitions involving hyperbolic inverse functions call for if a>u or u>a. I know that dealing with a definite integral we just use the limits of integration to figure that out, but what if we are dealing with an indefinite integral? How do you know then?