∫(01)x(arctanx)^2dx

Question

∫(0 1)x(arctanx)^2dx

Answer 1

∫[0→1] xarctan²x dx
=(1/2)∫[0→1] arctan²x d(x²)
=(1/2)x²arctan²x - ∫[0→1] x²arctanx/(1+x²) dx
=(1/2)x²arctan²x - ∫[0→1] (x²+1-1)arctanx/(1+x²) dx
=(1/2)x²arctan²x - ∫[0→1] arctanx dx + ∫[0→1] arctanx/(1+x²) dx
中间那个积分用分部积分,第三个积分直接凑微分
=(1/2)x²arctan²x - xarctanx + ∫[0→1] x/(1+x²) dx + ∫[0→1] arctanx d(arctanx)
=(1/2)x²arctan²x - xarctanx + (1/2)∫[0→1] 1/(1+x²) d(x²) + (1/2)arctan²x
=(1/2)x²arctan²x - xarctanx + (1/2)ln(1+x²) + (1/2)arctan²x |[0→1]
=(1/2)(π/4)² - π/4 + (1/2)ln2 + (1/2)(π/4)²
=π²/16 - π/4 + (1/2)ln2
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