Integrate $\int_0^\infty \frac{x\mathrm dx}{(x^2+a^2)(x^2+b^2)}$ 
Integrate $$\int_0^\infty \frac{x\mathrm dx}{(x^2+a^2)(x^2+b^2)}$$

Let, $x^2=\dfrac{1}{z}$ and, $2x\mathrm dx=-\dfrac{1}{z^2}\mathrm dz$
$$-\frac{1}{2}\int_\infty^0 \frac{\mathrm dz}{2+z(a^2+b^2)}$$
$$-\frac{1}{2(a^2+b^2)}\int_\infty^0 \frac{\mathrm dz}{\sqrt{\dfrac{4}{(a^2+b^2)^2}}+(\sqrt{z})^2}$$
$$-\frac{1}{2(a^2+b^2)}[\ln|\sqrt{z}+\sqrt{z+\frac{4}{(a^2+b^2)^2}}|]_\infty^0$$
But, I don't get the answer which my book got. I don't think that I have made any mistake. The answer of my book is $\frac{1}{a^2-b^2}\ln\frac{a}{b}$ If whole process of my work needed than, I will add it.

As regards your substitution $x^2=\dfrac{1}{z}$, you should have
$$\int_0^\infty \frac{x}{(x^2+a^2)(x^2+b^2)}\,dx=-\frac{1}{2}\int_\infty^0 \frac{dz}{(1+za^2)(1+zb^2)}.$$
Assuming that $a^2\not=b^2$, I propose a simpler substitution $t=x^2$:
$$\begin{align}\int_0^\infty \frac{x}{(x^2+a^2)(x^2+b^2)}\,dx&=
\frac{1}{2}\int_0^\infty \frac{1}{(t+a^2)(t+b^2)}\,dt\\
&=\frac{1}{2(a^2-b^2)}\int_0^\infty \left(\frac{1}{t+b^2}-\frac{1}{t+a^2}\right)\,dt\\
&=\frac{1}{2(a^2-b^2)}\left[\ln\left(\frac{t+b^2}{t+a^2}\right)\right]_0^\infty=\frac{\ln(|a/b|)}{a^2-b^2}.
\end{align}$$
