Question

求如下定积分详细解答步骤 最好有多种方法谢谢~

Answer 1

∫[3/4，4/3] √(θ²+1)/θ² dθ
令θ=tanβ，dθ=sec²βdβ
θ=3/4，β=arctan(3/4)、θ=4/3，β=arctan(4/3)
原式= ∫secβ/tan²β * sec²β dβ
= ∫secβcsc²β dβ
= ∫secβ(1+cot²β) dβ
= ∫secβ dβ + ∫cotβcscβ dβ
= ln|secβ+tanβ| - cscβ
= ln(5/3+4/3)-ln(5/4+3/4) - (5/4 - 5/3)
= ln(3/2) + 5/12

Answer 2

=积分{从根[1+(3/4)^2]到根[1+(4/3)^2]}x^2/(x^2-1)dx=根[1+(4/3)^2]-根[1+(3/4)^2]+0.5ln{《根[1+(4/3)^2]-1》/《根[1+(4/3)^2]+1》}-0.5ln{《根[1+(3/4)^2]-1》/《根[1+(3/4)^2]+1》}