Question

求x^4/((1+x^2)^(5/2))的不定积分

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Answer 1

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Answer 2

x = tany、dx = sec²y dy
siny = x/√(1 + x²)、cosy = 1/√(1 + x²)、secy = √(1 + x²)
∫ x⁴/(1 + x²)^(5/2) dx
= ∫ tan⁴y/sec⁵y · (sec²y dy)
= ∫ tan⁴ycos³y dy
= ∫ sin⁴y/cosy dy
= ∫ (1 - cos²y)²/cosy dy
= ∫ (1 - 2cos²y + cos⁴y)/cosy dy
= ∫ secy dy - 2∫ cosy dy + ∫ cos³y dy
= ∫ secy dy - 2∫ cosy dy + ∫ (1 - sin²y) d(siny)
= ln|secy + tany| - 2siny + siny - (1/3)sin³y + C
= ln|x + √(1 + x²)| - x/√(1 + x²) - x³/[3(1 + x²)^(3/2)] + C