# Evaluate $\int_{1}^{\sqrt{2}} \frac{x^4}{(x^2-1)^2+1}\,dx$

Evaluate the integral: $$\int_{1}^{\sqrt{2}} \frac{x^4}{(x^2-1)^2+1}\,dx$$

The denominator is irreducible, if I want to factorize and use partial fractions, it has to be in complex numbers and then as an indefinite integral, we get $$x + \frac{\tan^{-1}\left(\displaystyle\frac{x}{\sqrt{-1 - i}}\right)}{\sqrt{-1 - i}} + \frac{\tan^{-1}\left(\displaystyle\frac{x}{\sqrt{-1 + i}}\right)}{\sqrt{-1 + i}}+C$$

But evaluating this from $$1$$ to $$\sqrt{2}$$ is another mess, keeping in mind the principal values. I also tried the substitution $$x \mapsto \sqrt{x+1}$$, which then becomes

$$\frac{1}{2}\int_{0}^1 \frac{(x+1)^{3/2}}{x^2+1}\,dx$$

I don't see where I can go from here. Another substitution of $$x\mapsto \tan x$$ also leads me nowhere.

Should I approach the problem in some other way?

• WolframAlpha seems to do much the same thing you did, but the closed-form solution doesn't look pretty. – saulspatz Jul 12 '20 at 22:31
• The denominator has degree four. No polynomial of degree four is irreducible over the real numbers. You can factor it as a product of two quadratics without real coefficients. – Gerry Myerson Jul 13 '20 at 3:13
• Sorry, two quadratics with real coefficients. Have you tried it? – Gerry Myerson Jul 14 '20 at 12:28
• Are you still here, reggie? Have you tried factoring the quartic into two real quadratics? – Gerry Myerson Jul 15 '20 at 13:52
• @GerryMyerson Hi, sorry for such a late response. Thank you for your idea. It does factors into real quadratics. Although, as Quanto showed, it was not necessary to do the factoring. – reggie Jul 15 '20 at 15:18

Note \begin{align} I=&\int_{1}^{\sqrt{2}} \frac{x^4}{(x^2-1)^2+1}\,dx\\ = &\int_{1}^{\sqrt{2}} \left(1+\frac{2x^2-2}{x^4-2x^2+2}\right)\,dx\\ = &\sqrt2-1+\int_{1}^{\sqrt{2}} \frac{2-\frac2{x^2}}{x^2+\frac2{x^2}-2}dx\\ =& \sqrt2-1 + (1+\frac1{\sqrt2})I_1 + (1-\frac1{\sqrt2})I_2\tag1\\ \end{align}

where

\begin{align} I_1= \int_{1}^{\sqrt{2}} \frac{1-\frac{\sqrt2}{x^2}}{x^2+\frac2{x^2}-2}dx &=\int_{1}^{\sqrt{2}} \frac{d(1+\frac{\sqrt2}{x})}{(x+\frac{\sqrt2}x)^2-2(1+\sqrt2)}=0 \\ I_2= \int_{1}^{\sqrt{2}} \frac{1+\frac{\sqrt2}{x^2}}{x^2+\frac2{x^2}-2}dx &=\int_{1}^{\sqrt{2}} \frac{d(1-\frac{\sqrt2}{x})}{(x-\frac{\sqrt2}x)^2+2(\sqrt2-1)}\\ &=\sqrt{\frac2{\sqrt2-1}} \tan^{-1}\sqrt{\frac{\sqrt2-1}2} \end{align}

Plug $$I_1$$ and $$I_2$$ into (1) to obtain

$$I = \sqrt2-1 + \sqrt{\sqrt2-1}\tan^{-1}\sqrt{\frac{\sqrt2-1}2}$$

Start by writing $$x^4 = (x^2 - 1 + 1)^2$$ $$\implies x^4 = (x^2-1)^2 + 1 + 2(x^2-1)$$

So the our integral becomes:

$$\int_1^{\sqrt2}\frac{(x^2-1)^2 + 1 + 2(x^2-1)}{(x^2-1)^2 + 1}\,dx$$ $$= \int_1^{\sqrt2}\frac{(x^2-1)^2 + 1}{(x^2-1)^2 + 1}\,dx + 2\int_1^{\sqrt2}\frac{(x^2-1)}{(x^2-1)^2 + 1}\,dx$$ $$= \sqrt2 - 1 + 2\int_1^{\sqrt2}\frac{(x^2-\sqrt2 + 1-\sqrt2)}{(x^4 - 2x^2 + 2)}\,dx$$ $$= \sqrt2 - 1 + 2\int_1^{\sqrt2}\frac{(1-\sqrt2/x^2 )}{((x + \sqrt2/x)^2 - 2 - 2\sqrt2)}\,dx + 2(1-\sqrt2)\int_1^{\sqrt2}\frac{1}{(x^2-1)^2 + 1}\,dx$$ Here, I've split the integral so I can use the fact that after dividing the numerator and denominator by $$x^2$$, I can complete a square (The square of $$(x + \sqrt2/x))$$ and I'll have its derivative in the numerator for an easy substitution.

Put $$(x + \sqrt2/x) \rightarrow t$$ in the first integral, you can see that the upper and lower limits become the same $$(1+ \sqrt2)$$ So the first integral becomes $$0$$ and you're left with: $$\sqrt2 - 1 + 2(1-\sqrt2)\int_1^{\sqrt2}\frac{1}{(x^2-1)^2 + 1}\,dx$$ I was trying to avoid using complex numbers, but this integral becomes so much easier if you write: $$(x^2-1)^2 + 1 = (x^2 - 1 + i)(x^2 - 1 + i)$$ and use partial fractions. $$=\sqrt2 - 1 + \frac{1 - \sqrt2}{i}\int_1^{\sqrt2}\left(\frac{1}{x^2-1-i} + \frac{1}{x^2-1+i}\right)\,dx$$

$$=\sqrt2 - 1 + \frac{1 - \sqrt2}{i}\left(\frac{\tan^{-1}\left(\displaystyle\frac{x}{\sqrt{-1 - i}}\right)}{\sqrt{-1 - i}} + \frac{\tan^{-1}\left(\displaystyle\frac{x}{\sqrt{-1 + i}}\right)}{\sqrt{-1 + i}}\right)\Bigg|_{x=1}^{x=\sqrt2}$$