How to solve integrals where you can't factor a polynomial?

Question

Hi there guys I don't know if the title of the question should be the one for this but the thing is that I'm trying to solve this integral $\int \frac {\frac 12-u^2}{2u^4-2u^2+1}$$du$ and I have this doubt where if I try to do this $2(u^2)^2-2(u)^2+1$ in the denominator, I can't see how to factor it, plus I've had completed the square from what I've already tried to do a trigonometric substitution but seems that is not the best way to go since it gets way complicated, I see the term "irreducible factorization" from wolfram alpha when I try to factor it from there but that's something I don't know about yet, so it is possible to do this without that "irreducible factorization"? If so, what do I need to know? I do know the ways of integration but when I find problems like this one where you can't factor a polynomial like this one I get stuck really bad, I hope this question isn't too silly or something I'm just really trying to learn this real bad and I don't know somewhere else to ask!

Answer 1

I hope that you will be given a simpler way. Meanwhile, I give you my approach.

For the time being, let me set $u^2=x$ which makes the integrand $$\frac{1-2 x}{2 \left(2 x^2-2 x+1\right)}$$ Solving $2x^2-2x+1=0$ makes $$2(2x^2-2x+1)=4(x-a)(x-b)$$ with $a=\frac{1-i}2$ and $b=\frac{1+i}2$. Now, partial fraction decomposition makes $$\frac{1-2 x}{2 \left(2 x^2-2 x+1\right)}=\frac{1-2 x}{4(x-a)(x-b)}=\frac{1-2 a}{4 (a-b) (x-a)}+\frac{2 b-1}{4 (a-b) (x-b)}$$ that is to say $$\frac{1-2 x}{2 \left(2 x^2-2 x+1\right)}=\frac A {x-a}+\frac B {x-b}$$ where, after simplifications $A=B=-\frac 14$.

Now, back to $u$, we then have $$ \frac {\frac 12-u^2}{2u^4-2u^2+1}=-\frac 14\Big(\frac 1 {u^2-a}+\frac 1 {u^2-b}\Big)$$ and partial fraction decomposition can again be done.

So, by the end, you will just need to integrate four terms $\frac 1 {u+\alpha_i}$ where the $\alpha_i$'s are complex numbers.

I am sure that you can take it from here.

Edit

If you give the integral to Wolfram Alpha, it will give $$I=\int\frac {\frac 12-u^2}{2u^4-2u^2+1}\,du=-\frac{\tan ^{-1}\left(\frac{u}{\sqrt{-\frac{1}{2}-\frac{i}{2}}}\right)}{2 \sqrt{-2-2 i}}-\frac{\tan ^{-1}\left(\frac{u}{\sqrt{-\frac{1}{2}+\frac{i}{2}}}\right)}{2 \sqrt{-2+2 i}}$$ whci is just a linear combination of logarithms with complex terms since it rewrite $$I=\frac{1}{4} \sqrt{1+i} \left(\tanh ^{-1}\left(u\sqrt{1+i} \right)-(-1)^{3/4} \tanh ^{-1}\left(u\sqrt{1-i} \right)\right)$$ with $$\sqrt{1\pm i}=\sqrt[4]{2} \Big(\cos \left(\frac{\pi }{8}\right)\pm i \sin \left(\frac{\pi }{8}\right)\Big)$$

Answer 2

In a comment, you wondered if the problem could be instead $$\int \frac {\frac 12-u^2}{2u^4-2u^2-1}\, du$$ Doing the same as before, we have $$2x^2-2x-1=2(x-a)(x-b)$$ with $$ \quad a=\frac{1}{2} \left(1-\sqrt{3}\right)<0\quad,\quad b=\frac{1}{2} \left(1+\sqrt{3}\right)>0$$ and, again, partial fraction decomposition makes$$\frac{1-2 x}{2 \left(2 x^2-2 x-1\right)}=\frac A {x-a}+\frac B {x-b}$$ with $A=\frac 14$, $B=-\frac 14$.

Back to $u$, $$\frac {\frac 12-u^2}{2u^4-2u^2-1}=\frac 14\Big(\frac 1 {u^2-a}-\frac 1 {u^2-b}\Big)$$ Now, we face simple integrals and the final result is simply $$I=\frac{1}{4} \left(\sqrt{\sqrt{3}-1} \tanh ^{-1}\left(u\sqrt{\sqrt{3}-1} u\right)-\sqrt{1+\sqrt{3}} \tan ^{-1}\left(u\sqrt{1+\sqrt{3}} \right)\right)$$

Taking into account your remark in a comment, there is probably one more typo in the textbook.