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.

• Another change: consider $t=\sqrt{x^2-49}$.
– A.Γ.
Commented May 18, 2020 at 4:54
• math.stackexchange.com/questions/1606590/… Commented May 18, 2020 at 4:58
• Yet another substitution: $t = 7 \sec \theta$. I think the reason that OP uses hyperbolic functions is because the question says so. Commented May 18, 2020 at 4:59

\begin{align} &\ \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}(\frac\pi2- \text{arctan} \frac7{\sqrt{x^2-49}})\\ =&\ - \frac{1}{7}\text{arctan} \frac7{\sqrt{x^2-49}}+C \end{align}
My advice is to make the substitution $$x=7 \sec u$$ in the original quesion. Then all that remains is to integrate $$\frac{1}{7}.$$
$$\cosh^2 t - \sinh^2 t =1$$ so $$\sinh t = \sqrt{\cosh^2 t- 1}$$.
Then refer back to the substitution $$x = 7 \cosh t$$.
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