Integrating a function from negative infinity to infinity i got this weird question about integration from infinity to infinity.
$$\int_{-\infty}^{\infty} \frac{1}{z^2+25}$$
First idea was to take the factor out to get $(z+5)(z-5)$ but that really did not achieve anything.
Then i tried let $x= z^2$  and the $\frac{dx}{dz}= 2z$ then dz = 1/2z but that also went nowhere
Then i tried..... $x=z$, $\frac{dx}{dz}= 1$ hence $dz = dx$ 
$$\int_{-\infty}^{\infty} \frac{dx}{x^2+25}$$ which has not achieved anything. Please help!
Wait i just remember how this applies to partial fractions. But still confused how to grom infinity to infinity.
So start wit
$$\frac{1}{z^2+25}=\frac{1}{(z+5)(z-5)}$$
$$\frac{1}{(z+5)(z-5)}=\frac{A}{x+5}+\frac{B}{x-5}$$
$$1=A(x-5)+B(x+5)$$
At x = 5 $1=10B$ hence $B=0.1 
At x=-5 $A=1/-10$
But i'm not sure where to go from there

I think i finally got it
let $z=5tan(x)$ hence $dz=5sec^2(x)dx$ sub this in to the integrand
$$\int_{-\infty}^{\infty} \frac{dz}{z^2+25} =\int_{-\infty}^{\infty} \frac{5sec^2(x)dx}{25tan^2(x)+25}$$
as sec^2(x)=1+tan^2(x)
$$\int_{-\infty}^{\infty} \frac{5(tan^2x+1)}{25(tan^2(x)+1)}= \int_{-\infty}^{\infty} \frac{1dx}{5}$$
$$\int_{-\infty}^{\infty} \frac{dz}{z^2+25} = \frac{x}{5}$$ for  z =-$\infty,\infty$
as $$z = 5tan(x)$$ hence $$x=arctan(z/5)$$
observing its graph we find that as x approach positive and negative infinity artctan(z) equal $\frac{\pi}{2},-\frac{\pi}{2}$
this is the same for $x=arctan(z/5)$
hence, $$\int_{-\infty}^{\infty} \frac{dz}{z^2+25} =\frac{1}{5}(\frac{\pi}{2}-(-\frac{\pi}{2}) = \frac{\pi}{5}$$
