I'm asked to integrate $\int{dx\over x^2 - 9}$ using hyperbolic substitution. Using the relation cosh^2-1 = sinh^2, I let x = 3cosh(u), and through simplification, arrived at $\int$ ${1\over 3 csch(u)}du$. I integrated and got $1\over 3$ln(tanh(${u \over 2}$)+c. This is where I'm stuck, as I don't know how to convert back into terms of 'x' from 'u'. If anyone could help me out, it would be much appreciated!
Thank you!
On
The most straightforward way to go is to apply inverse hyperbolic functions (similar to arcsine and friends). Then \begin{align} x &= 3 \cosh u \\ \frac{x}{3} &= \cosh u \\ \cosh^{-1} \left( \frac{x}{3} \right) &= u \text{.} \end{align} Analogous to the inverse (circular) trigonometric functions, one should remember that this inverse function has a restricted domain, $[1,\infty)$, which can cause similar surprises.
Then it's useful to know the alternative forms of the arc-hyperbolic functions. For arc hyperbolic cosine, you want $$ \cosh^{-1} \alpha = \ln(\alpha + \sqrt{\alpha^2 - 1}) $$
best method is to use partial fractions $$\begin{align} \int\frac{dx}{x^2-9}dx\\=\frac{1}{6}\int\left(\frac{1}{x-3}-\frac{1}{x+3}\right)dx\\ =\frac{1}{6}(\ln|x-3|-\ln|x+3|)\\ =\frac{1}{6}\ln|(x-3)/(x+3)|+C \end{align}$$