Question

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

Answer 1

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