Compute $\int\frac {x^2}{x^4+1}dx$ via partial fractions

Question

I am trying to solve it with "partial fractions"

$$\frac {x^2}{x^4+1}=\frac{x^2}{(x^2+x\sqrt{2}+1)(x^2-x\sqrt{2}+1)}=\frac{Ax+B}{(x^2+x\sqrt{2}+1)}+\frac{Cx+D}{(x^2-x\sqrt{2}+1)}$$

and I get the following system of equations:

$A+C=0$

$-\sqrt{2}A+B+\sqrt{2}C+D=1$

$A-\sqrt{2}B+C+\sqrt{2}D=0$

$B+D=0$

How can I find $A,B,C,D$?

Answer 1

Observe that the integrand is an even fraction, so it is invariant when we change $x$ to $-x$: $$\frac{Ax+B}{(x^2+x\sqrt{2}+1)}+\frac{Cx+D}{(x^2-x\sqrt{2}+1)}=\frac{-Ax+B}{(x^2-x\sqrt{2}+1)}+\frac{-Cx+D}{(x^2+x\sqrt{2}+1)}.$$ This implies that $\;C=-A,\; D=B$, and you have only a linear system in two unknowns.

Answer 2

From $A+C=0$ and $A-B\sqrt2+C+D\sqrt2=0$ we have $$-B\sqrt2+D\sqrt2=0$$ which with $B+D=0$, gives $B=D=0$.

Now from $A+C=0$ and $-A\sqrt2+B+C\sqrt2+D=1$, noting that $B=D=0$, we have $$A=-C\implies C\sqrt2+C\sqrt{2}=1\implies C=\frac{1}{2\sqrt2}$$ and $$A=-\frac{1}{2\sqrt2}$$

Answer 3

You can solve the (easy) $4\times4$ system of linear equations, but notice that

$$(x^2+\sqrt2 x+1)-(x^2-\sqrt2 x+1)=2\sqrt2x$$ so that

$$\frac x{2\sqrt2}\frac{(x^2+\sqrt2 x+1)-(x^2-\sqrt2 x+1)}{(x^2+\sqrt2 x+1)(x^2-\sqrt2 x+1)}$$ does the trick.

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Notice that this will work for any positive power of $x$ at the numerator, hence any polynomial. Anyway, the constant term requires an extra twist:

$$(x^2+\sqrt2 x+1)+(x^2-\sqrt2 x+1)=2x^2+2,$$ from which you can cancel out $2x^2$.