Question

高数定积分题目,如果太麻烦可以只说思路.

Answer 1

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