solving an integral with trig substitution

[studentMCCS]

Hi. I'm struggling with this one, and can't tell if I'm getting the correct answers.

integral(5/(25x^2-9)^(1/2),x)

I tried two ways.

First way:

=integral (5/((225/9)x^2-(225/15))^(1/2),x)
=integral (1/(x^2-1/15)^(1/2),x)

x =(1/15)^(1/2)sec(u) dx=(1/15)^(1/2)sec(u)tan(u)du

x^2-1/15=(1/15)sec^2(u)-(1/15)
=(1/15)tan^2(u)

original integral = integral ( (1/15)^(1/2)sec(u)tan(u)/((1/15)tan^2(u))^(1/2),u)
=integral(sec(u),u)
=ln |sec(u)+tan(u)| + c

=ln | x/(1/5)^(1/2)+(15x^2-1)^(1/2) | +c

answer I get with my calculator is : ln((25x^2-9)^(1/2)+5x)

I think this I got it wrong, maybe it's right just in another form?

So I tried another way, but I'm not sure I did it right.

integral(5/(25x^2-9)^(1/2),x)

5x=3sec(u) dx=(3/5)sec(u)tan(u)du
25x^2-9=225tan^2(u)

original integral = integral ((5*(3/5)sec(u)tan(u)du)/(225tan^2(u)^(1/2),u)
=integral ((1/5)sec(u),u)

=(1/5) ln |sec(u)+tan(u) | +c

and so I made a triangle for reference

hypotenuse=5x
opposite=(25x^2-9)^(1/2)
adjacent=3

but I'm not sure this is correct?

And from that I got that the answer is : (1/5) ln |5x/3+(25x^2-9)^(1/2)/3| + c

Can anyone help me figure out what I did wrong, or how to get my answer into the form the calculator is giving?

Thanks

Student MCCS

 

[studentMCCS]

I have found my main errors. Here are a few photos of my work. I still am not getting the result my calculator is getting, but cannot figure out why. Sorry for the mess I made in the first post.