Question

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

Answer 1

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

Divide the numerator and the denominator by $x^2$

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

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

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

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

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

$=\frac{1}{2}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+2}dx-\frac{1}{2}\int \frac{d(x+\frac{1}{x})}{{(x-\frac{1}{x})}^2-2}dx$

Here we have to invoke a couple of standard integrals

$\Rightarrow \int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)$ and $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}ln\left|\frac{x-a}{x+a}\right|$

Using these are we can write the integrals obtained in last step as

$=\frac{1}{2}.\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)-\frac{1}{2}.\frac{1}{2\sqrt{2}}ln\left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|+c$

$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{x\sqrt{2}}\right)-\frac{1}{4\sqrt{2}}ln\left|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|+c$

Hence, solved.