We will use integration by parts. Assume that $a$ is positive. We first make the preliminary substitution $x=at$. Then $dx=a\,dt$, and when we go through the substitution process, we end up with $\int \frac{1}{a^3}\frac{dt}{(1+t^2)^2}$.
Let's not bother to carry the constant $\frac{1}{a^3}$ around, it can be inserted at the end. So we go after $\int \frac{dt}{(1+t^2)^2}$.
We use a little trick that has a number of uses. Let $I=\int \frac{dt}{1+t^2}$. (This is not a typo!) We recognize this integral instantly, since we know that the derivative of $\arctan t$ is $\frac{1}{1+t^2}$. But let's begin to evaluate the integral by using integration by parts.
Let $u=\frac{1}{1+t^2}$ and $dv=dt$. Then $du= -\frac{2t}{(1+t^2)^2}$ and we can take $v=t$. Thus $I=\int \frac{dt}{1+t^2}=\frac{t}{1+t^2}- \int-\frac{2t^2\,dt}{(1+t^2)^2}.$ Get rid of the doubled minus signs, and rewrite $2t^2$ as $2+2t^2 -2$. We end up with $I=\frac{t}{1+t^2} +\int \frac{2(1+t^2)}{(1+t^2)^2}-\int \frac{2\,dt}{(1+t^2)^2},$ and therefore $I=\frac{t}{1+t^2}+2I-2\int\frac{dt}{(1+t^2)^2}.$ Thus $2\int\frac{dt}{(1+t^2)^2}=\frac{t}{1+t^2}+I=\frac{t}{1+t^2}+\arctan t.$ So to find $\int \frac{dt}{(1+t^2)^2}$, divide the right-hand side by $2$. And don't forget to add the arbitrary constant $C$ of integration at the end.