<s>[INST] Integrate $\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$ 
$$\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$$

What I did was trig substitution:
$$x=\frac 13 \sec \theta$$ $$dx=\frac 13 \sec \theta \tan \theta \, d\theta$$
Then my integral becomes $$\int_{\frac 13 \sec\frac{\sqrt{2}}{3}}^{\frac 13 \sec\frac 23} \frac{81}{\sec^4 \theta} \, d\theta$$
But I got stuck here. What can I do next?
 [/INST] Substitute $y=3x$
$$I= \int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{1}{x^5\sqrt{9x^2-1}}dx = 81\int^2_{\sqrt2} \frac{1}{y^5\sqrt{y^2-1}}dy
$$
and then apply the reduction formula
$$\int \frac{1}{y^n\sqrt{y^2-1}}dy=I_n=\frac{\sqrt{y^2-1}}{(n-1)y^{n-1}}+\frac{n-2}{n-1}I_{n-2}
$$
to obtain
$$I=81\left( \frac{7\sqrt3-16}{64} +\frac38 \int^2_{\sqrt2} \frac{1}{y\sqrt{y^2-1}}dy\right)=\frac{81}{64}(7\sqrt3-16+2\pi)
$$
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