超难的不定积分,不信你试试!∫1/﹙1+x· tanx﹚²dx
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/08/01 11:53:50
超难的不定积分,不信你试试!∫1/﹙1+x· tanx﹚²dx
![超难的不定积分,不信你试试!∫1/﹙1+x· tanx﹚²dx](/uploads/image/z/15247830-30-0.jpg?t=%E8%B6%85%E9%9A%BE%E7%9A%84%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%2C%E4%B8%8D%E4%BF%A1%E4%BD%A0%E8%AF%95%E8%AF%95%21%E2%88%AB1%EF%BC%8F%EF%B9%991%2Bx%C2%B7+tanx%EF%B9%9A%26%23178%3Bdx)
设原函数 = A/(1 + xtanx) + C
[(1 + xtanx)A' - A(tanx + xsec²x)]/(1 + xtanx)²
令(1 + xtanx)A' - A(tanx + xsec²x) = 1
若A = tanx,则A' = sec²x
(1 + xtanx)(sec²x) - tanx(tanx + xsec²x)
= sec²x + xtanxsec²x - tan²x - xtanxsec²x
= sec²x - tan²x
= 1
所以∫ dx/(1 + xtanx)² = tanx/(1 + xtanx) + C
∫ dx/(1 + xtanx)
= ∫ [tanx(1 + xtanx) + (x - xtan²x - tanx)]/[x(1 + xtanx)²] dx
= ∫ tanx/[x(1 + xtanx)²] + ∫ (x - xtan²x - tanx)/[x(1 + xtanx)²] dx,分部积分法
= tanx/(1 + xtanx) - ∫ x * (x - xtan²x - tanx)/[x²(1 + xtanx)²] dx + ∫ (x - xtan²x - tanx)/[x(1 + xtanx)²] dx
= tanx/(1 + xtanx) + C
[(1 + xtanx)A' - A(tanx + xsec²x)]/(1 + xtanx)²
令(1 + xtanx)A' - A(tanx + xsec²x) = 1
若A = tanx,则A' = sec²x
(1 + xtanx)(sec²x) - tanx(tanx + xsec²x)
= sec²x + xtanxsec²x - tan²x - xtanxsec²x
= sec²x - tan²x
= 1
所以∫ dx/(1 + xtanx)² = tanx/(1 + xtanx) + C
∫ dx/(1 + xtanx)
= ∫ [tanx(1 + xtanx) + (x - xtan²x - tanx)]/[x(1 + xtanx)²] dx
= ∫ tanx/[x(1 + xtanx)²] + ∫ (x - xtan²x - tanx)/[x(1 + xtanx)²] dx,分部积分法
= tanx/(1 + xtanx) - ∫ x * (x - xtan²x - tanx)/[x²(1 + xtanx)²] dx + ∫ (x - xtan²x - tanx)/[x(1 + xtanx)²] dx
= tanx/(1 + xtanx) + C