Question

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

Answer 1

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