0

$$\int \frac{x^2+1}{x^4+1} dx$$

How can I solve this integral? Can the denominator be factored because I don't have difference of squares (because of the plus sign) ?

If I make the denominator $x^2*x^2+1$ I cannot cross out the terms $x^2+1$ so what do I need to do?

Thank you.

J. M. ain't a mathematician
  • 75,051
Harvey Ross
  • 11

4 Answers4

4

Factoring isn't hard: $$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt{2}x)^2=(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1),$$ so $$\frac{x^2+1}{x^4+1}=\frac12\left(\frac1{x^2+\sqrt{2}x+1}+\frac1{x^2-\sqrt{2}x+1}\right),$$ because $$\frac1{x^2+\sqrt{2}x+1}+\frac1{x^2-\sqrt{2}x+1}=\frac{(x^2-\sqrt{2}x+1)+(x^2+\sqrt{2}x+1)}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}=\frac{2(x^2+1)}{x^4+1}.$$ That means $$\int\frac{x^2+1}{x^4+1}\,dx=\frac12\left(\int\frac1{x^2+\sqrt{2}x+1}\,dx+\int\frac1{x^2-\sqrt{2}x+1}\,dx\right).$$

  • I am sorry, I didn't quite understand that. Where did the 1/2 in front of the parentheses come from? Also, does the (x^2+1)^2 from the numerator go into the equation you posted? Is the integral completely transformed into this equation? – Harvey Ross Aug 06 '17 at 17:12
1

Divide the numerator and denominator by $x^2$ and you get $\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}$ which equals $$\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+(\sqrt{2})^2}$$

Now put $x-\frac{1}{x}=t$ and $dt$ becomes $(1+\frac{1}{x^2})dx$.

Vidyanshu Mishra
  • 10,143
0

Divide by $x^2$. Put $x~-1/x $ as $t $

Upstart
  • 2,613
0

HINT:$$ x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt{2}x)^2=...$$ for your first integral

Dr. Sonnhard Graubner
  • 95,283