Question

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

Answer 1

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