Question

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

Answer 1

Let us consider the integral

$I_1=\int \frac{x^2+y}{x^4+x}dx$

$I_2=\int \frac{x^2-1}{x^4+1}dx$

$I_1=\int \frac{x^2(1+\frac{1}{x^2})}{x^2(x^2+\frac{1}{x^2})}dx=\int \frac{(1+\frac{1}{x^2})}{(x^2+\frac{1}{x^2})}dx$

let the substitute $\Rightarrow x-\frac{1}{x}=u$ so that $(1+\frac{1}{x^2})dx=du$

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

$I_1=\int \frac{1}{u^2+2}du=\frac{1}{\sqrt{2}}$ are $tan\left(\frac{u}{\sqrt{2}}\right)$

As $u=x-\frac{1}{x}=\frac{x^2-1}{x}$

$I_1=\frac{1}{\sqrt{2}}arctan\left(\frac{x^2-1}{\sqrt{2}x}\right)...\left(1\right)$

$I_2=\int \frac{x^2-1}{x^4+1}=\int \frac{x^2(1-\frac{1}{x^2})}{x^2(x^2+\frac{1}{x^2})}dx=\int \frac{x^2-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$

take $x+\frac{1}{x}=v$ so that $(1-\frac{1}{x^2})dx=du$

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

$I_2=\int \frac{1}{v^2-(\sqrt{2}{)}^2}dv=\frac{1}{2\sqrt{2}}ln\left|\frac{v-\sqrt{2}}{v+\sqrt{2}}\right|$

Replacing v by $x+\frac{1}{x}=\frac{x^2+1}{x}$

$I_2=\frac{1}{2\sqrt{2}}ln|\frac{x^2-2\sqrt{2}x+1}{x^2-\sqrt{2}x+1}|...\left(2\right)$

$I=\int \frac{1}{x^4+1}dx=\frac{1}{2}\int \frac{(x^2+1)-(x^2-1)}{x^4+1}dx=\frac{1}{2}{I}_1-\frac{1}{2}{I}_2$

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