TITLE: Integration of $1/(x\sqrt{25x^2-1})$
QUESTION [1 upvotes]: $$\int{\frac{1}{x\sqrt{25x^2-1}}}\,dx$$
Let $x=\frac{1}{5}u$
Now when I substite it and simplify I get
$$\int{\frac{1}{u\sqrt{u^2-1}}}\, du$$
There is a trig identity which says that this is equal to $$\sec^{-1} u$$
But this is incorrect.  I can't figure this out.  Where did I go wrong?

REPLY [1 votes]: $$\int\frac{dx}{x\sqrt{25x^2-1}}=\int\frac{25x}{25x^2\sqrt{25x^2-1}}dx$$
Let $\sqrt{25x^2-1}=u\implies\dfrac{50x\ dx}{2\sqrt{25x^2-1}}=du$ and $25x^2-1=u^2\iff25x^2=u^2+1$
{"set_name": "stack_exchange", "score": 1, "question_id": 1067310}