definite integrals with trigonometric substitution How do you integrate $$\int_{-7}^{-5} \frac{2}{x^4\sqrt{x^2-25}}dx,$$ I seem to be getting my answer off by -1 all the time. 

Note that since the integrand is an even function, we can write
$$\int_{-7}^{-5}\frac{2}{x^4\sqrt{x^2-25}}\,dx=\int_5^7\frac{2}{x^4\sqrt{x^2-25}}\,dx$$
Now, enforcing the substitution $x\to 5\sec(x)$ reveals 
$$\begin{align}
\int_5^7\frac{2}{x^4\sqrt{x^2-25}}\,dx&=\int_{0}^{\arccos(5/7)}\frac{2}{3125\sec^4(x)\tan(x)}\,5\sec(x)\tan(x)\,dx\\\\
&=\frac2{625}\int_0^{\arccos(5/7)}\cos^3(x)\,dx\\\\
&=\frac2{625}\int_0^{\arccos(5/7)}\left(\frac34 \cos(x)+\frac14\cos(3x)\right)\,dx\\\\
&=\frac3{1250}\sin(\arccos(5/7))+\frac{1}{3750}\sin(3\arccos(5/7))\\\\
&=\frac3{1250}\frac{2\sqrt{6}}{7}+\frac{1}{3750}\frac{102\sqrt{6}}{343}\\\\
&=\frac{164\sqrt{6}}{214375}
\end{align}$$
