Is there any other method to solve $\int \frac{x^2 - 1}{(x^2+1)\sqrt{x^4+1}} dx$

Question

Question is $\newcommand{\dd}{\,\mathrm{d}}$

$$\int \frac{x^2 - 1}{(x^2+1)\sqrt{x^4+1}} \dd x$$

Solution given in book $$\int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)\cdot\sqrt{x^2+\frac{1}{x^2}}}\dd x$$

$$\int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)\cdot\sqrt{\left(x+\frac{1}{x}\right)^2-2}}\dd x$$

Let $t=x+\frac{1}{x} $

$$\int \frac{1}{t\cdot\sqrt{t^2-2}}\dd t$$

$$\frac{1}{2} \mathrm{arcsec}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right) + C$$

If there was not any square root in the question then I could have used partial fractions to solve this.

But now I cannot think of any other way to do this.

So is there any other solution to this problem?

Answer 1

“If there wasn’t a square root there...” is not proper justification for there being an easier way to do something. For example I could find an elementary antiderivative of $\sin(x^2)$ if only there wasn’t a sine. This is the most straightforward way, once you can use the $x\pm\frac1x$ trick then this integral is almost trivial. I’m sure there’s another way to do this, perhaps rationalize and then partial fraction $$\sqrt{x^4+1}\left(\frac{x^2-1}{(x^2+1)(x^4+1)}\right)$$ Or maybe do something awful like $x^2=\tan\theta$ or $x^2=\sinh\theta$. All of these ideas would be a jumping off point but antiderivatives are unique so they will be some form of what you already found.

Answer 2

This method is a valid alternative to the original derivation but is perhaps not quite as easy to motivate or stumble upon.

First, consider $u=\frac{x}{\sqrt{x^4+1}}$ Then $du=-\frac{x^4-1}{(x^4+1)^{3/2}}dx=-\frac{x^4-1}{(x^4+1)\sqrt{x^4+1}}dx$.

This substitution is admittedly rather difficult to think of, although with enough trial and error or experience one may eventually stumble onto it. A possible motivation would be the observation that the integrand $\frac{x^2 - 1}{(x^2+1)\sqrt{x^4+1}}$ resembles an application of the quotient rule to some rational function involving a factor of $\frac{1}{\sqrt{x^4+1}}$, which may lead one to try the above substitution.

Anyway, a little rearrangement yields

$$-\frac{x^4+1}{(x^2-1)(x^2+1)}du=\frac{1}{\sqrt{x^4+1}}dx$$

Substituting what we have so far, we have

$$I=\int \frac{x^2 - 1}{(x^2+1)\sqrt{x^4+1}}dx=-\int \frac{x^2 - 1}{x^2+1} \frac{x^4+1}{(x^2-1)(x^2+1)}du=-\int \frac{x^4+1}{(x^2+1)^2}du$$

The integrand at this point is still in terms of $x$, but the differential is $du$. We hence seek to express $\frac{x^4+1}{(x^2+1)^2}$ in terms of $u$.

Note that

$$u^2=\frac{x^2}{x^4+1}$$

$$2u^2+1=\frac{2x^2}{x^4+1}+1=\frac{x^4+1+2x^2}{x^4+1}=\frac{(x^2+1)^2}{x^4+1}$$

And thus

$$I=-\int \frac{x^4+1}{(x^2+1)^2}du=-\int \frac{du}{2u^2+1}=-\frac{1}{\sqrt 2} \arctan(\sqrt{2}u)+C=-\frac{1}{\sqrt 2}\arctan(\frac{x\sqrt 2}{\sqrt{x^4+1}})+C$$

Answer 3

Note that $(\frac x{1+x^2})’= \frac{1-x^2}{(1+x^2)^2}$ \begin{align}\int \frac{x^2 - 1}{(x^2+1)\sqrt{x^4+1}} \ dx =&\int \frac{\frac{x^2-1}{(x^2+1)^2}}{\sqrt{\frac{x^4+1}{(x^2+1)^2}}}\ dx =\int \frac{-d(\frac x{1+x^2})}{\sqrt{1-\frac{2x^2}{(x^2+1)^2}}}\\ =& - \frac1{\sqrt2}\sin^{-1}\frac{\sqrt2 x}{x^2+1}+C \end{align}