Question

求解不定积分题目，题目如下图，有三道，非常感谢！

Answer 1

令x = sinz，dx = cosz dz
∫ x/√(1 - x²) · e^(arcsinx) dx
= ∫ sinz/cosz · e^z · cosz dz
= ∫ e^z · sinz dz = ∫ e^z d(- cosz)
= - (e^z)cosz + ∫ cosz de^z = - (e^z)cosz + ∫ e^z d(sinz)
= - (e^z)cosz + (e^z)sinz - ∫ sinz de^z
= - (e^z)cosz + (e^z)sinz - ∫ e^z · sinz dz
2∫ e^z · sinz dz = (e^z)(sinz - cosz)
∫ e^z · sinz dz = (1/2)(e^z)(sinz - cosz) + C
==> ∫ x/√(1 - x²) · e^(arcsinx) dx = (1/2)(e^arcsinx)(x - √(1 - x²) + C

令v = x^(1/6)，x = v⁶，dx = 6v⁵ dv
∫ √x/[x^(2/3) - √x] dx
= ∫ v³/(v⁴ - v³) · 6v⁵ dv
= 6∫ v⁵/(v - 1) dv
= 6∫ [v⁴ + v³ + v² + v + 1/(v - 1) + 1] dv
= 6[(1/5)v⁵ + (1/4)v⁴ + (1/3)v³ + (1/2)v² + ln|v - 1| + v| + C
= (6/5)x^(5/6) + (3/2)x^(2/3) + 2√x + 3x^(1/3) + 6^(1/6) + 6ln|1 - x^(1/6)| + C

令x = tanv，dx = sec²v dv，sinv = x/√(1 + x²)
∫ 1/[x⁴√(x² + 1)] dx
= ∫ sec²v/(tan⁴vsecv) dv
= ∫ cos³v/sin⁴v dv
= ∫ cot³vcscv dv
= ∫ cotvcscv · (csc²v - 1) dv
= ∫ (csc²v - 1) d(- cscv)
= cscv - (1/3)csc³v + C
= [(2x² - 1)√(x² + 1)]/(3x³) + C