Differentiation uner the integral sign - help me find my mistake This is my integral:
$$I(a)=\int_0^\infty\frac {\ln(a^2+x^2)}{(b^2+x^2)}dx.$$
Taking the first derivative with respect to a:
$$I'(a)=\int_0^\infty \frac {2adx} {(a^2+x^2)(b^2+x^2)}.$$
This is how I did the partial fraction decomposition:
$\frac {2a} {(a^2+x^2)(b^2+x^2)}=\frac {Ax+B} {(a^2+x^2)}+\frac {Cx+D} {(b^2+x^2)}$.
From here I get that $A=C=0$, $B=\frac {2a} {(b^2-a^2)}$ and $D=\frac {-2a} {(b^2-a^2)}$
Is this correct? Because when I try to solve $I'(a)$ using these values for $B$ and $D$ I get a different solution from the textbook?

Let’s redo the work$$\frac 1{(a^2+x^2)(b^2+x^2)}=\frac {Ax+B}{a^2+x^2}+\frac {Cx+D}{b^2+x^2}$$Multiplying both sides by the common denominator$$1=(Ax+B)(b^2+x^2)+(Cx+D)(a^2+x^2)$$To find values for $A$, $B$, $C$, and $D$, first set $x^2=-a^2$. Thus$$\begin{align*}1 & =(Ax+B)(b^2-a^2)\\ & =Ax(b^2-a^2)+B(b^2-a^2)\end{align*}$$Therefore, it’s easy to see that $A=0$ and $B=1/(b^2-a^2)$. Now do a similar procedure for the other term by setting $x^2=-b^2$. Thus$$\begin{align*}1 & =(Cx+D)(a^2-b^2)\\ & =Cx(a^2-b^2)+D(a^2-b^2)\end{align*}$$Thus, $C=0$ and $D=1/(a^2-b^2)$. So to sum everything up$$\frac 1{(a^2+x^2)(b^2+x^2)}\color{blue}{=\frac 1{(b^2-a^2)(a^2+x^2)}+\frac 1{(a^2-b^2)(b^2+x^2)}}$$
