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If f(x)=x∫1tan−1(t)t dt ∀x∈R+, and the value of f(e2)−f(1e2)=kπ2, then k=

Answer» If f(x)=x1tan1(t)t dt xR+, and the value of f(e2)f(1e2)=kπ2, then k=


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