高数定积分题目,如果太麻烦可以只说思路.
1个回答
K = ∫(0->1) ln(1+x) / (1+x²) dx
Let x = tanQ,dx = sec²QdQ
K = ∫(0->π/4) ln(1+tanQ) / (1+tan²Q) * sec²Q dQ
= ∫(0->π/4) ln(1+tanQ) dQ
Let Q = π/4 - y and dQ = -dy
K = ∫(π/4->0) ln[1+tan(π/4-y)] (-dy)
= ∫(0->π/4) ln{1 + [tan(π/4)-tany]/[1+tan(π/4)tany]} dy
= ∫(0->π/4) ln[1 + (1-tany)/(1+tany)] dy
= ∫(0->π/4) ln[(1+tany+1-tany)/(1+tany)] dy
= ∫(0->π/4) ln[2/(1+tany)] dy
= ∫(0->π/4) [ln2 - ln(1+tany)] dy
2K = ∫(0->π/4) ln2 dy = ln2 * (π/4)
K = (πln2)/8
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