How to calculate $\frac{1}{2\pi i}\int_{|z|=3}\frac{e^{z t}}{(z^2+1)^2}$?

Question

I tried to find the partial fractions of $\frac{1}{(z^2+1)^2}$ then

$\frac{1}{(z^2+1)^2}=\frac{A}{(z+i)^2}+\frac{B}{(z+i)}+\frac{C}{(z-i)^2}+\frac{D}{(z-i)}$

$1=A(z-i)^2+B(z+i)(z-i)^{2}+C(z+i)^2+D(z+i)^2(z-i)$

I got $A=D=-\frac{1}{4}$ but I am not sure how to get $B$ and $C$?

I there a way to find the values of $B$ and $C$? Also, is there a way to solve this integral without finding the partial fractions?

Answer 1

Hint:

Letting $\;C_1,\,C_2\;$ being little circle around the poles $\;i,\,-i\;$ resp., we have that

$$\frac1{2\pi i}\oint_{|z|=3}\frac{e^{zt}}{(z^2+1)^2}dz=\frac1{2\pi i}\left[\oint_{C_1}\frac{\frac{e^{zt}}{(z+i)^2}}{(z-i)^2}dz+\oint_{C_2}\frac{\frac{e^{zt}}{(z-i)^2}}{(z+i)^2}dz\right]=$$

$$=\left.\left(\frac{e^{zt}}{(z+i)^2}\right)'\right|_{z=i}+\left.\left(\frac{e^{zt}}{(z-i)^2}\right)'\right|_{z=-i}$$