Is there a way to solve $\int \frac1{(x^2+a^2)^2}$ without $x=\tan t$ substitution?

Question

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 ?

Answer 1

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.

Answer 2

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

$$ I=\frac{1}{8} \left[ (\frac{x}{x^2+4})+\int \frac{1}{x^2+4} dx \right] $$