<s>[INST] Indefinite intergral of $\int { \sqrt{ x^2-a^2} \over x } \mathrm dx $ I need to integrate $$\int { \sqrt{x^2-a^2} \over x } \mathrm dx $$ using substition, and show it equals $$ \sqrt{x^2 - a^2} - a(\operatorname{arcsec} ({x\over a })) +c $$
I've tried $x=a\sin t$ but I couldn't finish it out. 
Thank in advance for any help. 
 [/INST] Calculating this integral can be done in this way:
$$\int { \sqrt{x^2-a^2} \over x } dx =\int { x^2-a^2 \over x\sqrt{x^2-a^2} } dx= \int { x \over \sqrt{x^2-a^2} } dx-\int { a^2 \over x\sqrt{x^2-a^2} } dx =\sqrt{x^2-a^2}(+-)a\cdot\int {(\frac{x}{a})'\over \sqrt{1-(\frac{x}{a})^2)} } dx=\sqrt{x^2-a^2}(+-)a\cdot\arcsin(\frac{x}{a})+C.$$
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