Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{x^4+x^2+1} \\ =\frac{1}{2}\int \frac{2dx}{x^4+x^2+1} \\ \Rightarrow \frac{1}{2}\int \left(\frac{\left(x^2+1\right)-\left(x^2-1\right)}{x^4+x^2+1}\right)dx \\ \Rightarrow \frac{1}{2}\int \left(\frac{x^2+1}{x^4+x^2+1}\right)dx-\frac{1}{2}\int \left(\frac{x^2-1}{x^4+x^2+1}\right)dx \\ \mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}{x}^2 \\ I=\frac{1}{2}\int \left(\frac{1+{\displaystyle \frac{1}{x^2}}}{x^2+1+{\displaystyle \frac{1}{x^2}}}\right)dx-\frac{1}{2}\int \left(\frac{1-{\displaystyle \frac{1}{x^2}}}{x^2+1+{\displaystyle \frac{1}{x^2}}}\right)dx \\ =\frac{1}{2}\int \left(\frac{1+{\displaystyle \frac{1}{x^2}}}{x^2+{\displaystyle \frac{1}{x^2}}-2+3}\right)dx-\frac{1}{2}\int \frac{\left(1-{\displaystyle \frac{1}{x^2}}\right)dx}{x^2+{\displaystyle \frac{1}{x^2}}+2-1} \\ =\frac{1}{2}\int \frac{\left(1+{\displaystyle \frac{1}{x^2}}\right)dx}{{\left(x-{\displaystyle \frac{1}{x}}\right)}^2+{\left(\sqrt{3}\right)}^2}-\frac{1}{2}\int \frac{\left(1-{\displaystyle \frac{1}{x^2}}\right)dx}{{\left(x+{\displaystyle \frac{1}{x}}\right)}^2-1^2} \\ \mathrm{Putting}x-\frac{1}{x}=t \\ \Rightarrow \left(1+\frac{1}{x^2}\right)dx=dt \\ \mathrm{Putting}x+\frac{1}{x}=p \\ \Rightarrow \left(1-\frac{1}{x^2}\right)dx=dp \\ \therefore I=\frac{1}{2}\int \frac{dt}{t^2+{\left(\sqrt{3}\right)}^2}-\frac{1}{2}\int \frac{dp}{p^2-1^2} \\ =\frac{1}{2}\times \frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{t}{\sqrt{3}}\right)-\frac{1}{2}\times \frac{1}{2\times 1}\log \left|\frac{p-1}{p+1}\right|+C \\ =\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{x-{\displaystyle \frac{1}{x}}}{\sqrt{3}}\right)-\frac{1}{4}\log \left|\frac{x+{\displaystyle \frac{1}{x}}-1}{x+{\displaystyle \frac{1}{x}}+1}\right|+C \\ =\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right)-\frac{1}{4}\log \left|\frac{x^2-x+1}{x^2+x+1}\right|+C \end{gathered}$$