(sec2x −1)tan2x = 1\frac{\left(sec^2x\:-1\right)}{tan^2x}\:=\:1tan2x(sec2x−1)=1
−93⋅31-93\cdot31−93⋅31
∫x5(x6−9)15dx\int x^5\left(x^6-9\right)^{15}dx∫x5(x6−9)15dx
3+5∣3−5∣3+5\left|3-5\right|3+5∣3−5∣
tan2x+5=sec2x+4\tan^2x+5=\sec^2x+4tan2x+5=sec2x+4
7x−8⋅6x37x^{-8}\cdot6x^37x−8⋅6x3
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