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

Question

I came across a question today...

Find $$\displaystyle\int\dfrac{x^2-1}{(x^2+1)\sqrt{x^4+1}}\,dx$$

How to do this? I tried to take $x^4+1=u^2$ but no result. Then I tried to take $x^2+1=\frac{1}{u}$, but even that didn't work. Then I manipulated it to $\int \dfrac{1}{\sqrt{1+x^4}}\,dx+\int\dfrac{2}{(x^2+1)\sqrt{1+x^4}}\,dx$, butI have no idea how to solve it.

Wolframalpha gives some imaginary result...but the answer is $\dfrac{1}{\sqrt2}\arccos\dfrac{x\sqrt2}{x^2+1}+C$

Nikunj · Accepted Answer · 2016-03-31T12:14:38.400

11

Hint Divide the numerator and denominator by $x^2$, to get: $$\int \frac{(1-\frac{1}{x^2})dx}{(x+\frac{1}{x})(\sqrt {(x+\frac{1}{x})^2-2} )}$$ Then put $x+\frac{1}{x}=t$ $$\int \frac{dt}{t(\sqrt{t^2-2})}$$ Which is easily taken care of by putting $t=\sqrt2 \sec\theta$, $$\int \frac{(\sqrt2 \sec\theta\tan \theta)d\theta}{\sqrt2 \sec\theta \sqrt2 \tan\theta}$$

edited Mar 31 '16 at 12:14

answered Mar 02 '16 at 15:24

Nikunj

6,324

Very clever! +1 – zz20s Mar 02 '16 at 15:27
@zz20s Thanks!, btw someone was editing this, what was it?, I was adding something at that time! – Nikunj Mar 02 '16 at 15:29
I just removed some unnecessary brackets. If you'd prefer to keep them, that's okay. Either way, great answer. – zz20s Mar 02 '16 at 15:30
Thanks...today I found out that even WA can be wrong.. :) – manshu Mar 02 '16 at 15:32
@manshu I won't say that it can be wrong, could you post the result alpha gives? – Nikunj Mar 02 '16 at 15:33
Here – manshu Mar 02 '16 at 15:35
1

@manshu I take my words back – Nikunj Mar 02 '16 at 15:36
Setting $\sqrt{t^2-2}=u\implies\dfrac t{\sqrt{t^2-2}}dt=du, t^2=u^2+2$ $$\int\dfrac{dt}{t\sqrt{t^2-2}}=\int\dfrac{t\ dt}{t^2\sqrt{t^2-2}}=\int\dfrac{du}{u^2+2}=\dfrac1{\sqrt2}\arctan u$$ – lab bhattacharjee Mar 02 '16 at 15:41
@labbhattacharjee *$\frac{1}{\sqrt{2}} \tan^{-1}\frac{u}{\sqrt{2}}$ – Nikunj Mar 02 '16 at 15:48
@Nikunj, True & right – lab bhattacharjee Mar 02 '16 at 15:49

score 4 · Answer 2 · edited Apr 13 '17 at 12:20

4

With (this seem to be some kind of substitution of the day) $$ t=\frac{x}{\sqrt{1+x^4}} $$ you get $$ \int\frac{1}{1+2t^2}\,dt. $$ I'm sure you can take it from here.

edited Apr 13 '17 at 12:20

Community

1

answered Mar 02 '16 at 16:09

mickep

20,102

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

2 Answers2