∫ 1/(9+4x^2)dx=?
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∫ 1/(9+4x^2)dx=?
∫ 1/(9+4x^2)dx=?
∫ 1/(9+4x^2)dx=?
∫ 1/(9 + 4x²) dx
= ∫ 1/[4(9/4 + x²)] dx
= (1/4)∫ 1/[(3/2)² + x²] dx
= (1/4) * 1/(3/2) * arctan[x/(3/2)] + C
= (1/6)arctan(2x/3) + C
或
令2x = 3tanz,z = arctan(2x/3),2 dx = 3 sec²z dz
∫ 1/(9 + 4x²) dx = ∫ 1/(9 + 9tan²z) (3/2 * sec²z dz)
= ∫ 1/(9sec²z) * (3/2 sec²z dz)
= (1/6)∫ dz = z/6 + C
= (1/6)arctan(2x/3) + C
1/2*1/3arttan2x/3 c
∫(1-x)/[√(9-4x^2)]dx=
∫ 1/(9+4x^2)dx=?
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∫cos x / ( 1 + (sinx)^2 ) dx = ∫x^3 / ( 1 + x^4 ) dx = ∫(sec x)^3 * tan x dx = ∫x^2 * e^(-2x) dx = ∫x * cos 2x dx = ∫(cos 2x)^2 dx = ∫(13x - 6) / ( x (x-2)(x+3) )dx = ∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx = 请把过程写出来哈.>< 旷
这个怎恶魔算∫cos x / ( 1 + (sinx)^2 ) dx = ∫x^3 / ( 1 + x^4 ) dx = ∫(sec x)^3 * tan x dx = ∫x^2 * e^(-2x) dx = ∫x * cos 2x dx = ∫(cos 2x)^2 dx = ∫(13x - 6) / ( x (x-2)(x+3) )dx = ∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx =
∫1-x/(√9-4x^2)dx,
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