Question

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

• A. $\frac{1}{2\sqrt{2}}\log |\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}|+c$
• B. $\frac{1}{\sqrt{2}}\log |\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}|+c$
• C. $\frac{1}{2\sqrt{2}}\log |\frac{x^2+\sqrt{2}x-1}{x^2+\sqrt{2}x+1}|+c$
• D. $\frac{1}{2\sqrt{2}}\log |\frac{x^2-\sqrt{2}x-1}{x^2-\sqrt{2}x+1}|+c$

Answer 1

The correct option is A $\frac{1}{2\sqrt{2}}\log |\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}|+c$

$I=\int \frac{x^2-1}{x^4+1}dx$
$=\int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$ (dividing$N^r$and$D^r$by$x^2$)
$=\int \frac{(1-\frac{1}{x^2})}{{(x+\frac{1}{x})}^2-{\left(\sqrt{2}\right)}^2}dx$
Substitute $u=x+\frac{1}{x}$
$du=(1-\frac{1}{x^2})dx$
So, $I=\int \frac{du}{u^2-{\left(\sqrt{2}\right)}^2}$
$I=\frac{1}{2\sqrt{2}}\log |\frac{u-\sqrt{2}}{u+\sqrt{2}}|+C$
$I=\frac{1}{2\sqrt{2}}\log |\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}|+C$