Integrate $\frac{1}{x \cdot \sqrt{x^2 + 1}}$ without trigonometric substitutions

Question

For my current ODE class I need to solve the integral

$$\int\limits_1^z \frac{1}{u \cdot \sqrt{u^2 + 1}} \,du$$

Every solution I've seen uses trigonometric substitution (using $x = \tan(t)$), but I don't know all the integration rules (besides $\sin$ and $\cos$) by heart. Therefore I wanted to ask, whether it is possible to solve this integral without the use of trigonometric substitution?

The indefinite solution

$$\ln(u) - \ln(\sqrt{u^2 + 1} + 1)$$

doesn't exactly scream for trigonometric functions.

Answer 1

Hint

$$\int_1^z \frac{1}{u \cdot \sqrt{u^2 + 1}} \,du=\int_1^z \frac{u}{u^2 \cdot \sqrt{u^2 + 1}} \,du$$

Set $v= \sqrt{u^2+1} , dv =\frac{u}{\sqrt{u^2+1}}du$.

$$\int_1^z \frac{u}{u^2 \cdot \sqrt{u^2 + 1}} =\int_{\sqrt{2}}^{\sqrt{z^2+1}} \frac{dv}{v^2-1} \,du$$

Answer 2

This can be an another hint, that is to take $$x=\tfrac{1}{t}.$$