Find the integral using an appropriate trigonometric substitution.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{1}{x^2\sqrt[\phantom{2}]{x^{2}25}}dx$

a$.-\frac{\sqrt[\phantom{2}]{x^{2}25}}{5x}C$

b$.\frac{\sqrt[\phantom{2}]{x^{2}25}}{5x}C$

c$.\frac{\sqrt[\phantom{2}]{x^{2}25}}{25x}C$

d$.-\frac{\sqrt[\phantom{2}]{x^{2}25}}{25x}C$