$$\int \frac{1}{x\sqrt{x^2-25}} \,dx$$
So I realized that there are multiple ways to solve the question above:
- I could use the inverse secant identity: $\int \frac{1}{x\sqrt{x^2-1}} \, dx$, giving the final result in $\operatorname{arcsec}$
- Or simply substituting $u = \sqrt{x^2-25}$, giving the final result in $\arctan$
- Other possible methods that gives the final result in $\arccos$
I tried using method #2 and got $$dx = \frac{du \sqrt{x^2-25}}{x}$$ Using this $dx$ value, I attempted to get the original equation in terms of $du$, but failed miserably. $$ =\int \frac{1}{xu} \times \frac{\sqrt{x^2-25}}{x} \, du $$ $$ = \int \frac{(x^2-25)^\frac{1}{2}}{x^2} \times \frac{1}{u} \, du $$ which would not eliminate the variable $x$. So I was wondering what I did wrong, or could do better.
Also, would it be simpler to use method #1 in this case?
Edit
So, after looking at the comments, I was able to get: $$\int \frac{1}{u^2+25} \, du$$ But I'm still lost on what should be the next step as this is still not integratable yet.