Question

∫secx2·(tanx2/1+tanx)+dx

Answer 1

∫[sec²x·(tan²x/(1+tanx)]dx
=∫{(1+tanx)(tanx-1)+1]/(1+tanx)}dtanx
=∫[tanx-1+1/(1+tanx)dtanx
=½tan²x-tanx+ln|1+tanx|+C

Answer 2

∫ sec²x/(1 + tanx) dx
= ∫ d(tanx)/(1 + tanx) = ∫ d(1 + tanx)/(1 + tanx)
= ln|1 + tanx| + C
∫ tan⁵tsec³t dt
一全数又没程革位科节拉验照。
= ∫ tan⁴tsec²t * (secttant dt)
= ∫ (tan²t)²sec²t d(sect)
下所现革设件统见角决任极，步界观千往属。
= ∫ (sec²t - 1)²sec²t d(sect)
= ∫ (sec⁴t - 2sec²t + 1)sec²t d(sect)
= ∫ sec⁶t - 2sec⁴t + sec²t d(sect)
= (1/7)sec⁷t - (2/5)sec⁵t + (1/3)sec³t + C
∫ x³/√(1 + x²) dx
= ∫ x[(1 + x²) - 1]/√(1 + x²) dx
= ∫ x√(1 + x²) dx - x/√(1 + x²) dx
= (1/2)∫ √(1 + x²) - 1/√(1 + x²) d(1 + x²)
= (1/2)[(2/3)(1 + x²)^(3/2) - 2√(1 + x²)] + C
= (1/3)(x² - 2)√(1 + x²) + C
∫ dx/[1 + (x + 1)^(1/3)]，z³ = x + 1，3z² dz = dx
= 3∫ z²/(1 + z) dz
= 3∫ z[(1 + z) - 1]/(1 + z) dz
= 3∫ z - z/(1 + z) dz
= 3∫ z dz - 3∫ [(1 + z) - 1]/(1 + z) dz
= 3∫ z dz - 3∫ dz + 3∫ dz/(1 + z)
= (3/2)z² - 3z + 3ln|1 + z| + C
= (3/2)(x + 1)^(2/3) - 3(x + 1)^(1/3) + 3ln|1 + (1 + x)^(1/3)| + C
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Answer 3

被积函数是sec²xtan²x/(1+tanx)吗
I=∫sec²xtan²xdx/(1+tanx)=∫tan²xd(tanx)/(1+tanx)
令u=tanx
I=∫u²du/(1+u)=∫(u-1+1/(1+u))du=u²/2-u+ln(1+u)+C=tan²x/2-tanx+ln(1+tanx)+C