Hello, machi4velli!

Your work is absolutely correct!

. . The book did something really *stupid!*

$\displaystyle \int\frac{6\,dx}{x\sqrt{25x^2 -1}}$

I used trigonometric substitution:

$\displaystyle x \:=\:\frac{}{5}\sec\theta\quad\Rightarrow\quad dx \:=\:\frac{1}{5}\sec\theta\tan\theta\,d\theta$

Substitute: .$\displaystyle \int\frac{6\left(\frac{1}{5}\sec\theta\tan\theta\, d\theta\right)}{\left(\frac{1}{5}\sec\theta\right) (\tan\theta)} \;=\;6\int d\theta$

After integrating, I got: .$\displaystyle 6\theta + C$, so I did this:

. . $\displaystyle x = \frac{1}{5}\sec\theta\quad\Rightarrow\quad 5x = \sec\theta\quad\Rightarrow\quad \theta \:=\:\text{arcsec}(5x)$

So, this gives: .$\displaystyle 6\text{ arcsec}(5x) + C$ for my solution. . **Correct!**

However, the real solution is: .$\displaystyle 6\text{ arctan}\left(\sqrt{25x^2-1}\right) + C$ **??**

Their answer is correct . . . but unforgivably stupid!

You had: .$\displaystyle \sec\theta = 5x$

Consider the right triangle: Code:

*
* |
* | ________
5x * | √25x² - 1
* |
* θ |
* - - - - - - - - *
1

Since $\displaystyle \sec\theta = \frac{hyp}{adj}$, we have: .$\displaystyle hyp = 5x,\:adj = 1$

Using Pythagorus, we find that: .$\displaystyle opp = \sqrt{25x^2 - 1}$

Since $\displaystyle \tan\theta = \frac{\sqrt{25x^2-1}}{1}$, it is true that $\displaystyle \theta \:=\:\arctan(\sqrt{25x^2 - 1})$

. . but it is a truly **silly** way to write the answer!

By their reasoning, all of these are also correct answers:

. . $\displaystyle \begin{array}{cccc}\theta &= &\arcsin\left(\frac{\sqrt{25x^2-1}}{5x}\right) \\

\theta & = & \arccos\left(\frac{1}{5x}\right) \\

\theta & = & \text{arccot}\left(\frac{1}{\sqrt{25x^2 - 1}}\right) \\

\theta & = & \text{arccsc}\left(\frac{5x}{\sqrt{25x^2-1}}\right)

\end{array}$

but who (in their right minds) would insist on using them?