Definite integral using $u$-substitution. Evaluate the definite integral:
$$\int_{1/3}^{\sqrt{2}/3} \dfrac{1}{x\sqrt{9x^2-1}}{dx}$$
So,
I am to use $u$-substitution, and immediately, it would appear that perhaps the integral may be some function of x in the form $\sqrt{x}, \ln{x},$ or possibly some inverse trigonometric function.
So I feel like the most progress I've made was in taking $u = 9x^2 -1 \implies du = 18x {\ dx}$. The concept of this method is very strange to me, as well (though seemingly I know how to apply it, in most cases), because, conceptually, taking and moving the $dx$ in place of the $du$ arithmetically seems a little bit like having $1+1 = 11$. Perhaps I just don't understand exactly what it's all implying?
But, aside from not really understanding why $u$-substitution works, I really don't know what to take for $u$ in order to evaluate this integral using $u$-substitution.
 A: Try substituting $u=\sqrt{9x^2-1}$.  Then $du=\frac{9x}{\sqrt{9x^2-1}}dx=\frac{9x^2}{x\sqrt{9x^2-1}}dx$
Notice that this implies $$\frac{dx}{x\sqrt{9x^2-1}}=\frac{du}{9x^2}=\frac{du}{1+u^2}$$
Thus, 
$$\int_{1/3}^{\sqrt{2}/3} \frac{dx}{x\sqrt{9x^2-1}}=\int_{0}^{1} \frac{du}{1+u^2}=\arctan(1)-\arctan(0)=\pi/4$$

NOTE:  Primer on Integration by Substitution
The fundamental theorem of calculus states that if 
$$F(x)=\int_a^x f(t) dt$$
then
$$F'(x) = f(x)$$
Now, let $G$ be defined as the integral (This is a substitution $t=g(u)$ and we will show that $F=G$ to complete the primer.)
$$G(x)= \int_{g^{-1}(a)}^{g^{-1}(x)}f(g(u))\frac{dg(u)}{du}du$$
From the chain rule we have 
$$\begin{align}
G'(x) &=G'(g^{-1}(x))\frac{dg^{-1}(x)}{dx}\\\\
&=\left(f(g(g^{-1}(x)))\frac{dg(g^{-1}(x))}{dg^{-1}(x)}\right)\frac{dg^{-1}(x)}{dx}\\\\
&=f(x) \frac{dx}{dg^{-1}(x)}\frac{dg^{-1}(x)}{dx}\\\\
&=f(x)
\end{align}$$
Thus, we have that $F'(x) = G'(x)$.  Since $F(a)=G(a)$, then $F(x) = G(x)$.
A: It's an Arcsecant. We can rewrite that integral as $$\int_{\frac{1}{3}}^{\frac{\sqrt{2}}{3}} \frac{1}{x\sqrt{(\frac{x}{\frac{1}{3}})^2-1}}$$
Which is $$\frac{1}{3}\sec^{-1}x |_{\frac{1}{3}}^{\frac{\sqrt{2}}{3}}$$
Hope that helps.
A: Hint: On one hand, $\big(9x^2-1\big)'=18x$. On the other hand, $\dfrac1{x\sqrt{9x^2-1}}=\dfrac x{x^2\sqrt{9x^2-1}}=$
$=\dfrac{\bigg[\dfrac{\big(18x\big)}{18}\bigg]}{\bigg[\dfrac{~\big(9x^2-1\big)+1}9\bigg]\cdot\sqrt{9x^2-1}~}~.~$ Can you take it from here ? ;-$)$
