<s>[INST] Integrate $\int \frac {dx}{x\sqrt{x^2-49}}\,$ I'm trying to integrate the following:
$$\int \frac {dx}{x\sqrt{x^2-49}}\,$$
using the substitution $x=7\cosh(t)$
This is as far as I've gotten:
$\int \frac {dx}{x\sqrt{x^2-49}}\,$ = $\int \frac {7\sinh(t)dt}{7\cosh(t)7\sinh(t)}\,$ = $\int \frac {dt}{7\cosh(t)}\,$ = $\int \frac {\cosh(t)dt}{7\cosh^2(t)}\,$ = $\int \frac {\cosh(t)dt}{7(1+\sinh^2(t))}\,$
Let $u=\sinh(t)$, $du=\cosh(t)dt$
$$\int \frac {\cosh(t)dt}{7(1+\sinh^2(t))}\,
=\int \frac {du}{7(1+u^2)}\,$$
$$=\frac {1}{7}\arctan(u)+C=\frac{1}{7} \arctan(\sinh(t))+C$$
This is as far as I have been able to get. Somehow from here I need to get to 
$$-\frac{1}{7} \arctan(\frac{7}{\sqrt{x^2-49}})+C$$
Can someone please show me how to finish this integration problem off? I would appreciate it so so much.
 [/INST] Note
$$\frac{1}{7} \arctan(\sinh t)
=\frac{1}{7} \arctan\sqrt{\cosh^2t -1}\\
=\frac{1}{7} \arctan\sqrt{\frac{x^2}{49}-1}
=\frac{1}{7} \arctan\frac{\sqrt{x^2-49}}7\\
= \frac{1}{7} \text{arccot } \frac7{\sqrt{x^2-49}}
 = \frac{1}{7}(\frac\pi2-  \text{arctan} \frac7{\sqrt{x^2-49}})\\
 =- \frac{1}{7}\text{arctan} \frac7{\sqrt{x^2-49}}+C
$$
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