math calc help please

Question

integral dx / √x √x-1

thank you

Answer 1

I'll assume the problem is:

Let u = √x which means x = u²

And it also means du/dx = 1/(2√x), cross multiplying we get dx = 2√x du or dx = 2u du

Making those substitutions we get:

∫1/(u√(u² - 1))•2u du and then simplifying, we get:

2∫1/√(u² - 1) du

Now do a trig substitution u = sec(θ) which means du = sec(θ)tan(θ) dθ:

2∫1/(√(sec²(θ) -1)) sec(θ)tan(θ) dθ

Then using tan²(θ) = sec²(θ) - 1:

2∫1/(√tan²(θ)) sec(θ)tan(θ) dθ

2∫1/(tan(θ)) sec(θ)tan(θ) dθ

2∫sec(θ) dθ

The antiderivative of sec(θ) is ln(tan(θ) + sec(θ)) so we get:

2ln|tan(θ) + sec(θ)| + C

Undo the u-substitution and the trig substitution to get:

2ln|√x + √(x - 1)| + C