Use Laplace transform (with respect to t)to calculate the integral
I=\int([\cos(tx)/(x^2+a^2)]dx t\geqslant 0
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Let
$\displaystyle I(t) =\int_{0}^{\infty}\frac{\cos(tx)}{x^2+a^2}dx$
Then
$\displaystyle \mathcal{L}\{I\} =\int_{0}^{\infty}\frac{\mathcal{L}\{\cos(tx)\}}{x ^2+a^2}dx=\int_{0}^{\infty}\frac{s}{s^2+x^2}\cdot \frac{1}{x^2+a^2}dx$
Now by partial factions we get
$\displaystyle \mathcal{L}\{I\} =\frac{s}{s^2-a^2}\int_{0}^{\infty}\left( \frac{1}{x^2+a^2}-\frac{1}{x^2+s^2}\right)dx=\frac{\pi}{2}\left( \frac{s}{a(s^2-a^2)}-\frac{1}{s^2-a^2}\right)$
Now if you take the inverse Laplace transform we get
$\displaystyle I=\frac{\pi}{2a}\cosh(at)-\frac{\pi}{2}\sinh(at)$
$\displaystyle \frac{s}{(x^2+a^2)(x^2+s^2)}=\frac{Ax+B}{x^2+a^2}+ \frac{Cx+D}{x^2+s^2}$
$\displaystyle s =(Ax+B)(x^2+s^2)+(Cx+D)(x^2+a^2)$
Now expand all of this out to get
$\displaystyle s=Ax^3+Bx^2+s^2Ax+s^2B + Cx^3+Dx^2+a^2Cx+a^2D $
$\displaystyle s=(A+C)x^3+(B+D)x^2+(s^2A+a^2C)x+(s^2B+a^2D)$
So this gives 4 equations in the 4 unknowns A,B,C,D
$\displaystyle A+C=0 \quad B+D=0 \quad s^2A+a^2C=0 \quad s^2B+a^2D=s$
dont forget that s and a are constants. Now just solve this system.