$\int (x^2+1)/(x^4+1)\ dx$

Question

I have Divided the numerator and Denominator by $x^2$ to get $\dfrac{1+x^{-2}}{x^2+x^{-2}}$ then changed it into $(1+(x^{-2}))/[(x-x^{-1})^2 +2]$ then took $x-(1/x)$ as $u$ and Differentiated it with respect to $x$ to get $dx=du/(1+x^{-2})$ Finally I got this expression:

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

After this I need help!

Just apply partial fraction decomposition by noticing that $$x^4+1 = (x^2+1)^2-(x\sqrt{2})^2 = (x^2+x\sqrt{2}+1)(x^2-x\sqrt{2}+1)$$ — Jack D'Aurizio, Jul 17 '17 at 18:24
Not sure what you mean by "then changed it into... But it should be $$x^2+x^{-2}=(x-x^{-1})^2+2$$You forgot the square. — Thomas Andrews, Jul 17 '17 at 18:49
So after setting $u=x-x^{-1}$, you get $\int \frac{du}{u^2+2}.$ Setting $v=u/\sqrt2$ gets you $\frac{1}{\sqrt{2}}\int \frac{dv}{1+v^2}=\frac{1}{\sqrt2}\arctan(v)+C.$ — Thomas Andrews, Jul 17 '17 at 18:54
Are you missing a square? Thus: $$ \frac{1+x^{-2}}{x^2+x^{-2}} = \frac {1+x^{-2}} {(x-x^{-1})^2 +2} $$ That seems to be clearly what you intended so I've edited the question accordingly. — Michael Hardy, Jul 17 '17 at 19:00
Amazing how much commentary is here without anyone mentioning that $\displaystyle \int (1+u^2)^{-1} ,du = \arctan u + \text{constant}. \qquad$ — Michael Hardy, Oct 07 '17 at 16:11

Michael Hardy · Answer 1 · 2017-10-07T17:05:55.353

7

Standard tables of integrals say: $$ \int (1+v^2)^{-1} \, dv = (\arctan v) + \text{constant}. $$

Therefore \begin{align} \int (2+u^2)^{-1} \, du & = \frac 1 {2\sqrt 2} \int \left(1 + \left(\frac u {\sqrt 2} \right)^2 \right)^{-1} \, \big( \sqrt 2\, du\big) \\[10pt] & = \frac 1 {2\sqrt 2} \int (1+v^2)^{-1} \, dv \\[10pt] & = \frac 1 {2\sqrt 2} \arctan v + C \\[15pt] & = \frac 1 {2\sqrt 2} \arctan \frac u {\sqrt 2} + C. \end{align}

edited Oct 07 '17 at 17:05

answered Jul 17 '17 at 18:57

Michael Hardy

1

I think it should be $\int(2+u^2)^{-1}du$, but it's beautiful! – Michael Rozenberg Jul 17 '17 at 19:11
Thanks Again Sir! I had this part done by the way then too. – user126358 Jul 17 '17 at 19:15

Michael Rozenberg · Answer 2 · 2017-07-17T18:28:32.793

6

Since $x^4+1=x^4+2x^2+1-2x^2=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$, we obtain $$\frac{x^2+1}{x^4+1}=\frac{1}{2}\left(\frac{1}{x^2+\sqrt2x+1}+\frac{1}{x^2-\sqrt2x+1}\right).$$ Now, use $\int\frac{1}{1+x^2}dx=\arctan{x}+C$.

edited Jul 17 '17 at 18:28

answered Jul 17 '17 at 18:27

Michael Rozenberg

203,855

Thanks Sir! Helped a lot – user126358 Jul 17 '17 at 18:28
@user126358 You are welcome! – Michael Rozenberg Jul 17 '17 at 18:29
2

@MichaelRozenberg : This fails to mention that what the original poster did was correct and actually gets you the answer more simply by going on from where the O.P. left off. (See what my posted answer.) – Michael Hardy Jul 17 '17 at 19:04
@MichaelHardy Thanks Sir! – user126358 Jul 17 '17 at 19:13

David G. Stork · Accepted Answer · 2017-07-17T22:15:02.573

0

Express MichaelRozenberg's denominator, $x^2 + \sqrt{2} x + 1$ in the form $u^2 + 1$ so you can use the $\tan^{-1}$ form.

Specifically:

$x^2 + \sqrt{2} x + 1$

Let $\sqrt{2}x + 1 = u$,

so

$u^2 = 2 x^2 + 2 \sqrt{2} x + 1$

so

$u^2 + 1 = 2 x^2 + 2 \sqrt{2} x + 2$

or

${1 \over u^2 + 1} = {1 \over x^2 + \sqrt{2} x + 1}$.

And likewise for the other term with the minus sign.

Now you have the arc tan integral formula, giving:

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

edited Jul 17 '17 at 22:15

answered Jul 17 '17 at 18:23

David G. Stork

30,409

Can u please tell me what to do After my last step? I m stuck – user126358 Jul 17 '17 at 18:26

$\int (x^2+1)/(x^4+1)\ dx$

3 Answers3