∫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

∫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
∫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.
>< 旷课2个月 算不出来了~
我晕倒我晕倒，你们算的我一个都没看明白，
自己啃书去。
国外和国内的差异竟然这么大。

∫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
∫cos x / ( 1 + (sinx)^2 ) dx =
= ∫d(sinx)/[1 + (sinx)^2] = arctan(sinx) + C
[arctan（）,反正切函数啊]
∫x^3 / ( 1 + x^4 ) dx =
=(1/4)∫d(1 + x^4)/(1 + x^4) = ln(1 + x^4)/4 + C
∫(sec x)^3 * tan x dx =
=∫(secx)^2d(secx)
= (secx)^3/3 + C
∫x^2 * e^(-2x) dx =
= (-1/2)∫x^2d(e^(-2x)) = -1/2e^(-2x)x^2 + ∫e^(-2x)xdx
= -1/2e^(-2x)x^2 - 1/2∫xd(e^(-2x))
= -1/2e^(-2x)x^2 - 1/2e^(-2x)x + ∫e^(-2x)dx
= -1/2e^(-2x)[x^2 + x + 1] + C
∫x * cos(2x) dx =
= (1/2)∫xdsin(2x) = 1/2xsin(2x) - 1/2∫sin(2x)dx
= 1/2xsin(2x) + 1/4cos(2x) + C
∫(cos 2x)^2 dx =
= (1/2)∫[1 + cos(4x)]dx
= 1/2[x + sin(4x)/4] + C
∫(13x - 6) / ( x (x-2)(x+3) )dx =
∫[1/x + 2/(x-2) + 3/(x+3)]dx = ln|x| + 2ln|x-2| + 3ln|x+3| + C
∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx =
= ∫[x^2 - x - 2 + 3x]dx/[(x-2)(x+1)]
= ∫[1 + 3x/[(x-2)(x+1)]]dx
= x + ∫[2/(x-2) + 1/(x+1)]dx
= x + 2ln|x-2| + ln|x+1| + C

你问这些 也太简单了 基础的 最上面的那个 把cosx拿到后面去∫1/ ( 1 + (sinx)^2 ) dsinx
然后套公式呗
太简单了 我觉定 你自己算吧

arctan就是tan的反函数

∫cos x/1+(sinx)^2)dx =∫1/1+(sinx)^2)dsinx =arctan(sinx)+C
∫x^3/(1 + x^4)dx =(1/4)∫1/(1 + x^4)dx^4=ln(1+x^4)+C
∫(sec x)^3*tanxdx =∫(sec x)^3dsecxx =(1/4)(sec x)^4+C
∫x^2*e^(-2x)dx=∫x^2*e^(-...

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∫cos x/1+(sinx)^2)dx =∫1/1+(sinx)^2)dsinx =arctan(sinx)+C
∫x^3/(1 + x^4)dx =(1/4)∫1/(1 + x^4)dx^4=ln(1+x^4)+C
∫(sec x)^3*tanxdx =∫(sec x)^3dsecxx =(1/4)(sec x)^4+C
∫x^2*e^(-2x)dx=∫x^2*e^(-2x)dx=-2x*e^(-2x)+∫x*e^(-2x)dx
=-2x*e^(-2x)+(-1/2)x*x*e^(-2x)+(1/2)∫e^(-2x)dx
=-2x*e^(-2x)+(-1/2)x*x*e^(-2x)-(1/4)e^(-2x)+C
∫x*cos2xdx =(1/2)x*sin2x-(1/2)∫cos2xdx
=(1/2)x*sin2x+(1/4)sin2x+C
∫(cos2x)^2dx=(1/2)∫(1+(cos4x))dx=(x/2)+(1/4)sin4x+C
∫(13x-6)/(x(x-2)(x+3))dx=∫((-29/30)+(1/x)+(2/5(x-2))+(1/(x+3))dx
=(-29/30)x+(lnx)+(2/5)ln(x-2)+ln(x+3)+C
∫(x^2+2x-2)/((x-2)(x+1))dx =∫(1+2/(x-2)+1/(x+1))dx
=x+2ln(x-2)+ln(x+1)+C

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∫cos x / ( 1 + (sinx)^2 ) dx = ∫1/( 1 + (sinx)^2 )d(sinx)=arctan(sinx)
∫x^3 / ( 1 + x^4 ) dx =1/4∫ 1/( 1 + x^4 ) d(x^4)
现在没空下次补

∫cos x / ( 1 + (sinx)^2 ) dx
=∫1/ ( 1 + (sinx)^2 )d(sin x)
=arctan(sin x) +C (C是常数)
∫x^3 / ( 1 + x^4 ) dx
=1/4*∫1/( 1 + x^4 ) d(x^4)
=1/4 * arctan(x^4)+C
∫(sec x)^3 * tan x...

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∫cos x / ( 1 + (sinx)^2 ) dx
=∫1/ ( 1 + (sinx)^2 )d(sin x)
=arctan(sin x) +C (C是常数)
∫x^3 / ( 1 + x^4 ) dx
=1/4*∫1/( 1 + x^4 ) d(x^4)
=1/4 * arctan(x^4)+C
∫(sec x)^3 * tan x dx
=∫(sec x)^2 d(sec x)
=1/3 * (sec x)^3 +C
∫x^2 * e^(-2x) dx
=-1/2 * ∫x^2 d [e^(-2x)]
=-1/2[x^2 * e^(-2x)-∫e^(-2x) d x^2]
=-1/2 * x^2 * e^(-2x)-1/2 * ∫x d [e^(-2x)]
=-1/2 * x^2 * e^(-2x)-1/2[x * e^(-2x)-∫e^(-2x) d x]
=-1/2 * x^2 * e^(-2x)-1/2 * x * e^(-2x)-1/4*e^(-2x) +C
∫x * cos 2x dx
=1/2*∫x d sin 2x
=1/2 * x * sin 2x-1/2*∫sin 2x dx
=1/2 * x * sin 2x+1/4* cos 2x +C
∫(cos 2x)^2 dx
= ∫(cos 4x + 1)/2 dx
=1/4 * ∫(cos 4x + 1)/2 d 4x
=1/8 * sin 4x +1/2 * x +C
∫(13x - 6) / ( x (x-2)(x+3) )dx
=∫{ 4/[x * (x+3)]+9/[x * （x-2)] } dx
=∫{4/3[1/x - 1/(x+3)]+9/2[1/(x-2) - 1/x]}dx
=-19/6 * ln x +9/2 * ln(x-2) -4/3 * ln(x+3) +C
∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx
=∫{1 + 3x/ [(x-2)(x+1)]} dx
=∫[1 + 2/ (x-2) +1/(x+1)]dx
=x + 2ln (x-2) + ln (x+1) +C
这样就搞定了，
不懂可以问我

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