Question

$\int \frac{x^2+1}{x^4+7x^2+1}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \left(\frac{x^2+1}{x^4+7x^2+1}\right)dx \\ \mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}{x}^2 \\ I=\int \left(\frac{1+{\displaystyle \frac{1}{x^2}}}{x^2+7+{\displaystyle \frac{1}{x^2}}}\right)dx \\ =\int \frac{\left(1+{\displaystyle \frac{1}{x^2}}\right)dx}{x^2+{\displaystyle \frac{1}{x^2}}-2+9} \\ \Rightarrow \int \frac{\left(1+{\displaystyle \frac{1}{x^2}}\right)dx}{{\left(x-{\displaystyle \frac{1}{x}}\right)}^2+3^2} \\ \mathrm{Putting}x-\frac{1}{x}=t \\ \Rightarrow \left(1+\frac{1}{x^2}\right)dx=dt \\ \therefore I=\int \frac{dt}{t^2+3^2} \\ =\frac{1}{3}{\tan }^{-1}\left(\frac{t}{3}\right)+\mathrm{C} \\ =\frac{1}{3}{\tan }^{-1}\left(\frac{x-{\displaystyle \frac{1}{x}}}{3}\right)+\mathrm{C} \\ =\frac{1}{3}{\tan }^{-1}\left(\frac{x^2-1}{3x}\right)+\mathrm{C} \end{gathered}$$