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Estas integrales no son nada fáciles, solo haré una por ejercicio.
$$\begin{align}&\int \frac{dx}{(a^2+x^2)^2}=\\&\\&x=a·tg\,t \implies t=arctg \frac xa\\&dx=a·sec^2t\\&\\&=\int \frac{a·sec^2t}{(a^2+a^2tg^2t)^2}dt=\\&\\&\int \frac{a·sec^2t}{[a^2(1+tg^2t)]^2}dt=\\&\\&\int \frac{a·sec^2t}{[a^2(sec^2t)]^2}dt=\\&\\&\int \frac{dt}{a^3sec^2t}=\int \frac{\cos^2t}{a^3}dt=\\&\\&\frac 1{a^3}\int \frac{1+\cos 2t}{2}dt=\\&\\&\frac{t}{2a^3}+\frac{sen2t}{4a^3}+C\\&\\&\frac{1}{2a^3}\left(arctg \frac xa +sen\left(2arctg \frac xa\right) \right)+C\\&\\&\text{y ahora viene lo que no gusta hacer}\\&\\&\frac{sen u}{cosu}=\frac xa\\&\\&senu = \frac xa cosu\\&\\&sen^2u=\frac {x^2}{a^2}\cos^2u\\&\\&sen^2u=\frac{x^2}{a^2}(1-sen^2u)\\&\\&sen^2u\left(1+\frac{x^2}{a^2} \right)=\frac{x^2}{a^2}\\&\\&sen^2u =\frac{x^2}{a^2+x^2}\\&\\&senu=\frac x{\sqrt{a^2+x^2}}\\&\\&Luego\\&\\&arctg \frac xa=arcsen \left(\frac x{\sqrt{a^2+x^2}}\right)\\&\\&\text{Y la integral queda}\\&\\&\frac{1}{2a^3}\left(arctg \frac xa +sen\left(2arcsen \left(\frac x{\sqrt{a^2+x^2}}\right)\right) \right)+C=\\&\\&\frac{1}{2a^3}\left(arctg \frac xa +2 \frac x{\sqrt{a^2+x^2}}\sqrt{1-\frac{x^2}{a^2+x^2}} \right)+C=\\&\\&\frac{1}{2a^3}\left(arctg \frac xa +2 \frac x{\sqrt{a^2+x^2}}\frac{a}{\sqrt{a^2+x^2}} \right)+C=\\&\\&\frac{1}{2a^3}\left(arctg \frac xa +\frac {2ax}{a^2+x^2} \right)+C\\&\\&\\&\end{align}$$
Y eso es todo.