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calc.- trig substitution

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s- integral

s 1/ [ (x^4) sq.rt(x^2+9)]

i know x=3tanx
sq.rt(x^2+9)= 3 secx
dx= 3/[cos^2(x)]

so far i know:
= 1/ (3tan^4(x)) 3secx cos^2(x)) dx
=1/ 81 [ (sin^4 (x)/cos^4 (x)) (1/cosx) (cos^2(x))]

then i'm not really sure what to do next

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