This is from an integral problems book, it is a part of the solution of a larger problem.
$$\int_0^1\frac1{(x^2+4)^2}~dx=\frac18\Biggl( \Bigl(\frac x{x^2+4} \Bigr) \bigg\vert _0^1~+~\int_0^1\frac{dx}{x^2+4} \Biggr ) = ~~ ... $$
It looks like integration by parts but it isn't.
I suppose this technique works generally for integrals of the type $ \frac1{(x^2+a^2)^2}$
What is this technique ?
In fact, more generally, you can compute $\displaystyle\int\dfrac{1}{(x^2+a^2)^{n+1}} {\rm d}x$ in terms of $\displaystyle\int\dfrac{1}{(x^2+a^2)^{n}} {\rm d}x$.
For, observe that $\dfrac{1}{(x^2+a^2)^{n}}=x\cdot\dfrac{x}{(x^2+a^2)^{n+1}}+\dfrac{a^2}{(x^2+a^2)^{n+1}}$. Now apply integration by part on the first term of the RHS.
Like this:
$$ I=\int \frac{1}{(x^2+4)^2}dx=\frac{1}{B} \int \frac{1}{2x} d(A-\frac{B}{x^2+4})=\frac{1}{B} \left[ \frac{1}{2x} (A-\frac{B}{x^2+4})+\int \frac{Ax^2+4A-B}{x^2+4} \frac{1}{2x^2} dx \right] $$
We note
$$ \frac{Ax^2+4A-B}{x^2+4} \frac{1}{2x^2} $$
We can let
$$ A=2,4A-B=0 \implies A=2,B=8 $$
$$ \frac{Ax^2+4A-B}{x^2+4} \frac{1}{2x^2}=\frac{1}{x^2+4} $$
Thus
$$ I=\frac{1}{8} \left[ (\frac{x}{x^2+4})+\int \frac{1}{x^2+4} dx \right] $$
