Question

求不定积分∫ (secx)^2/(1+tanx)dx ∫(tant)^5×(sect)^3dt ∫x^3/√(1+x^2)dx ∫dx/1+三次跟号项x+1

Answer 1

∫ sec²x/(1 + tanx) dx
= ∫ d(tanx)/(1 + tanx) = ∫ d(1 + tanx)/(1 + tanx)
= ln|1 + tanx| + C
∫ tan⁵tsec³t dt
= ∫ tan⁴tsec²t * (secttant dt)
= ∫ (tan²t)²sec²t d(sect)
= ∫ (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
有帮助，请采纳

Answer 2

∫ sec²x/(1 + tanx) dx
= ∫ d(tanx)/(1 + tanx) = ∫ d(1 + tanx)/(1 + tanx)
= ln|1 + tanx| + C

∫ tan⁵tsec³t dt
= ∫ tan⁴tsec²t * (secttant dt)
= ∫ (tan²t)²sec²t d(sect)
= ∫ (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