求x^3(1+4x^2)^(1/2)的不定积分
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求x^3(1+4x^2)^(1/2)的不定积分
如图
![](http://img.wesiedu.com/upload/e/38/e3861cdbc2c3d2c870ee668bb490dfe4.jpg)
如图
![](http://img.wesiedu.com/upload/e/38/e3861cdbc2c3d2c870ee668bb490dfe4.jpg)
![求x^3(1+4x^2)^(1/2)的不定积分](/uploads/image/z/17496495-63-5.jpg?t=%E6%B1%82x%5E3%281%2B4x%5E2%29%5E%281%2F2%29%E7%9A%84%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86)
换元法,很简单而已,令x = 1/2 · tanz,dx = 1/2 · sec²zdz
√(1 + 4x²) = secz
∫ x³√(1 + 4x²) dx
= ∫ (1/8)tan³z · secz · (1/2)sec²z dz
= (1/16)∫ tan²zsec²z d(secz)
= (1/16)∫ (sec²z - 1)sec²z d(secz)
= (1/16)∫ (sec⁴z - sec²z) d(secz)
= (1/16)(1/5 · sec⁵z - 1/3 · sec³z) + C
= (1/16)[1/5 · (1 + 4x²)^(5/2) - 1/3 · (1 + 4x²)^(3/2)] + C
= (1/120)(6x² - 1)(1 + 4x²)^(3/2) + C
√(1 + 4x²) = secz
∫ x³√(1 + 4x²) dx
= ∫ (1/8)tan³z · secz · (1/2)sec²z dz
= (1/16)∫ tan²zsec²z d(secz)
= (1/16)∫ (sec²z - 1)sec²z d(secz)
= (1/16)∫ (sec⁴z - sec²z) d(secz)
= (1/16)(1/5 · sec⁵z - 1/3 · sec³z) + C
= (1/16)[1/5 · (1 + 4x²)^(5/2) - 1/3 · (1 + 4x²)^(3/2)] + C
= (1/120)(6x² - 1)(1 + 4x²)^(3/2) + C