∫(0→1) arctan(e^x)/e^x dx

I'll give you some ideas. Can you figure it out for yourself
turn
arctan(e^x)/e^2x d(e^x)
Let e ^ x = u, then note that 0 → 1 becomes 1 → E
Then arctanu / U & # 178; Du = - arctanu D (1 / U)
Then the partial integral becomes - [arctanu / U - ∫ 1 / U D (arctanu)] = - [arctanu / U - ∫ 1 / u * 1 / (1 + U & # 178;)] Du
Now it's mainly about "1 / u * 1 / (1 + U & # 178;)
∫1/u*1/(1+u²)=∫[1/u-u/(1+u²)]du=lnu-1/2*ln(1+u²)
Then 1 → e is taken in
over.

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