Question

$\int \frac{x^2+9}{x^4+81}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \left(\frac{x^2+9}{x^4+81}\right)dx \\ \mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}{x}^2 \\ I=\int \frac{\left(1+{\displaystyle \frac{9}{x^2}}\right)dx}{x^2+{\displaystyle \frac{81}{x^2}}} \\ =\int \frac{\left(1+{\displaystyle \frac{9}{x^2}}\right)dx}{x^2+{\left({\displaystyle \frac{9}{x}}\right)}^2-2\times x\times {\displaystyle \frac{9}{x}}+2\times x\times {\displaystyle \frac{9}{x}}} \\ =\int \frac{\left(1+{\displaystyle \frac{9}{x^2}}\right)dx}{{\left(x-{\displaystyle \frac{9}{x}}\right)}^2+{\left(\sqrt{18}\right)}^2} \\ \mathrm{Putting}x-\frac{9}{x}=t \\ \Rightarrow \left(1+\frac{9}{x^2}\right)dx=dt \\ \therefore I=\int \frac{dt}{t^2+{\left(\sqrt{18}\right)}^2} \\ =\int \frac{dt}{t^2+{\left(3\sqrt{2}\right)}^2} \\ =\frac{1}{3\sqrt{2}}{\tan }^{-1}\left(\frac{t}{3\sqrt{2}}\right)+\mathrm{C} \\ =\frac{1}{3\sqrt{2}}{\tan }^{-1}\left(\frac{x-{\displaystyle \frac{9}{x}}}{3\sqrt{2}}\right)+\mathrm{C} \\ =\frac{1}{3\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-9}{3\sqrt{2}x}\right)+\mathrm{C} \end{gathered}$$