Evaluate the following integrals: x^2+9 x^4+81 dx

Question

Evaluate the following integrals:
$\int\frac{\text{x}^2+9}{\text{x}^4+81}\ \text{dx}$

✓

Answer

Let $\text{I}=\int\frac{\text{x}^2+9}{\text{x}^4+81}\ \text{dx}$Dividing numerator and denominator by $x^2$
$\text{I}=\int\frac{1+\frac{9}{\text{x}^2}}{\text{x}^2+\frac{81}{\text{x}^2}}\ \text{dx}$
$=\int\frac{1+\frac{9}{\text{x}^2}}{\Big(\text{x}-\frac{9}{\text{x}}\Big)^2+18}$
Let $\Big(\text{x}-\frac{9}{\text{x}}\Big)=\text{t}\Rightarrow\Big(1+\frac{9}{\text{x}^2}\Big)\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{\text{t}^2+18}$
$\text{I}=\frac{1}{3\sqrt{2}}\tan^{-1}\Big(\frac{\text{t}}{3\sqrt{2}}\Big)+\text{C}$
Thus,
$\text{I}=\frac{1}{3\sqrt{2}}\tan^{-1}\Big(\frac{\text{x}^2-9}{3\sqrt{2}}\Big)+\text{C}$

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