Evaluate using convolution theorem $$\int_0^\infty \frac{1}{(a^2+t^2)(b^2+t^2)} dt$$
I took $a,b > 0$ and consider a function $f(x)$ such that:
$$f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{e^{iwx}}{(a^2+w^2)(b^2+w^2)}$$
Required answer is $\pi f(0)$. Now, consider $\mathfrak{F}\{g(x)\} = \frac{1}{a^2 + w^2}$ and $\mathfrak{F}\{h(x)\} = \frac{1}{b^2+w^2}$, So we have:
$$f(x) = \frac{1}{2\pi}\int_{-\infty}^\infty \mathfrak{F}\{g(x)\}\mathfrak{F}\{h(x)\} e^{iwx} dw \\ = \frac{1}{2\pi}\int_{-\infty}^\infty \mathfrak{F}\{g *h\} e^{iwx}dw$$
or $f(x) = g * h$.
Now we know $\mathfrak{F}\{e^{-|ax|}\} = \frac{2|a|}{a^2 + w^2}$ so we just need to find $g * -h$ which is:
$$f(0) = \frac{1}{4ab}\int_{-\infty}^{\infty} e^{-|at|} e^{-|bt|} dt = \frac{1}{2ab(a+b)}$$
So $\pi f(0) = \frac{\pi}{2ab(a+b)}$???
What if we also did not know fourier inverse of $\frac{1}{a^2+w^2}$ ?
Your calculation is correct. We can double-check it with a somewhat easier approach, viz. $$\frac{1}{b^2-a^2}\int_0^\infty\left(\frac{1}{a^2+t^2}-\frac{1}{b^2+t^2}\right)dt=\frac{1}{b^2-a^2}\frac{\pi}{2}\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{\pi}{2ab(a+b)}$$in the case $a\ne b$ (the case $a=b$ can be handled by continuity using the dominated convergence theorem, but one advantage of Fourier convolution is not having to split the cases like that). But you ask a very good question: what if we don't know the Fourier inverse? I guess one has to either derive it or look it up. If $a>0$, everyone's favourite choice of semicircular contour gives $$\int_0^\infty\frac{e^{iwt}dt}{a^2+t^2}=\frac{i}{2a}\int_0^\infty\left(\frac{1}{t+ia}-\frac{1}{t-ia}\right)e^{iwt}dt=\frac{i}{2a}\left(0-2\pi i\exp -aw\right)=\frac{\pi}{a}\exp -aw.$$Thus for $a\ne 0$, the integral is $\frac{\pi}{|a|}\exp -|a|w$.