<s>[INST] Integrate $\int \frac {x^4}{\sqrt {x^2-9}} \,dx$ Integrate $\int \dfrac {x^4}{\sqrt {x^2-9}} dx$
My Attempt:
Let $x=3\sec (\theta )$
$$dx=3\sec (\theta).\tan (\theta).d\theta$$
Then,
$$=\int \dfrac {x^4}{\sqrt {x^2-9}}$$
$$=\int \dfrac {81. \sec^4 (\theta)}{\sqrt {(3\sec (\theta))^2 - 9}} 3\sec (\theta).\tan (\theta).d\theta $$
$$=\int \dfrac {81 \sec^5 (\theta). 3\tan (\theta).d\theta}{3\tan (\theta)}$$
$$=81\int \sec^5 (\theta) d\theta$$
 [/INST] Let $ x + t = \sqrt{x^2 - 9}$
$$I = \dfrac{-1}{16}\int \dfrac{(t^2 + 9)^4}{t^5}  dt = -\dfrac{486\ln(|t|)+\dfrac{t^4+72t^2}{4}-\dfrac{5832t^2+6561}{4t^4}}{16}+C$$
Then, $$\dfrac1{16}\left(\dfrac{t^4+72t^2}{4}-\dfrac{5832t^2+6561}{4t^4}\right) =- \left(\dfrac{27}8 x \sqrt{x^2 - 9} +  x^3\dfrac{1}4  \sqrt{x^2 - 9}\right) $$
$$I =  \dfrac{27}8 x \sqrt{x^2 - 9} +  x^3\dfrac{1}4  \sqrt{x^2 - 9} - \dfrac{243}8\ln(\sqrt{x^2 - 9} - x) + C$$
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