Seeking indefinite integral ∫ TaNx / x∧3

∫(tanx/x³)dx=(-1/2)∫tanx d(1/x²)=-(1/2)tanx*(1/x²)+(1/2)∫(1/x²)[1/(1+x²)]dx=-[(tanx)/(2x²)]+(1/2)∫dx/x² -(1/2)∫[1/(1+x²)]dx=-[(tanx)/(2x²)] -[1/(2x)] ...

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