Question

x^2/√1+4x^2积分怎么求？高数题

请问这道题怎么算需要详细方法

Answer 1

Answer 2

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

Answer 3

如图

Answer 4

1+x+x^2=3/4+(x+1/2)^2,换元,令x+1/2=3tany/4,这样就能去掉根号了（利用1+(tany)^2=1/(cosy)^2）.