Question

integration of dx / x⁴ +x² +1

Answer 1

$$\begin{gathered} \int \frac{dx}{x^4+x^2+1} \\ Or\int \frac{1}{2}\left(\frac{2-x^2+x^2}{x^4+x^2+1}\right)dx \\ Or\frac{1}{2}\int \frac{1-x^2}{x^4+x^2+1}dx+\frac{1}{2}\int \frac{1+x^2}{x^4+x^2+1}dx \\ Dividethenumeratoranddenominatorby{x}^2,wehave \\ \frac{1}{2}\int \frac{1/x^2-1}{x^2+1+1/x^2}dx+\frac{1}{2}\int \frac{1/x^2+1}{x^2+1+1/x^2}dx \\ Or\frac{1}{2}\int \frac{1+1/x^2}{x^2+1+1/x^2}dx-\frac{1}{2}\int \frac{1-1/x^2}{x^2+1+1/x^2}dx \\ Or\frac{1}{2}\int \frac{1+1/x^2}{(x-{\displaystyle \frac{1}{x}}{)}^2+(\sqrt{3}{)}^2}dx-\frac{1}{2}\int \frac{1-1/x^2}{(x+1/x{)}^2-(1{)}^2}dx \\ Let(x-1/x)=t,so(1+1/x^2)dx=dtandx+1/x=u,so(1-1/x^2)dx=du \\ So\frac{1}{2}\int \frac{dt}{(t{)}^2+(\sqrt{3}{)}^2}-\frac{1}{2}\int \frac{du}{(u{)}^2-(1{)}^2} \\ =\frac{1}{2}\times \frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{t}{\sqrt{3}}\right)-\frac{1}{2}\times \frac{1}{2}\log \left|\frac{u-1}{u+1}\right| \\ Or\frac{1}{2}\times \frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x-1/x}{\sqrt{3}}\right)-\frac{1}{2}\times \frac{1}{2}\log \left|\frac{x+1/x-1}{x+1/x+1}\right|+C \end{gathered}$$