integration of $\int {\frac{{dx}}{{(9 + 25x^2 )\sqrt {1 + x^2 } }} = \frac{1}{{12}}} \arctan (\frac{{4x}}{{3\sqrt {1 + x^2 } }}){\rm{ }}$

Question

we have with wolfram

$\int {\frac{{dx}}{{(9 + 25x^2 )\sqrt {1 + x^2 } }} = \frac{1}{{12}}} \arctan (\frac{{4x}}{{3\sqrt {1 + x^2 } }}){\rm{ }}$

we impose the change of variable

$x^{ - 1} = \sqrt {t^2 - 1} $

y get $ - \frac{1}{{12}}\arctan \left( {\frac{{3\sqrt {1 + x^2 } }}{{4x}}} \right)?$

Answer 1

$$\int {\frac{{dx}}{{(9 + 25x^2 )\sqrt {1 + x^2 } }} = \frac{1}{{12}}} \arctan (\frac{{4x}}{{3\sqrt {1 + x^2 } }}){\rm{ }}$$ and
$$\int {\frac{{dx}}{{(9 + 25x^2 )\sqrt {1 + x^2 } }} = - \frac{1}{{12}}\arctan \left( {\frac{{3\sqrt {1 + x^2 } }}{{4x}}} \right)} $$ is not surprising.

Note that the two results differ by a constant. This is due to the fact that if $$tan(\alpha )= \frac{{4x}}{{3\sqrt {1 + x^2 } }}{\rm{ }}$$ and $$tan(\beta)= - { {\frac{{3\sqrt {1 + x^2 } }}{{4x}}} } $$

Then $$ \alpha = \beta -\pi /2$$