三角函数积分∫arctan4t dt 怎么算?

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三角函数积分∫arctan4t dt 怎么算?
三角函数积分
∫arctan4t dt 怎么算?

三角函数积分∫arctan4t dt 怎么算?
让4x=t
4 dx = dt
dx = (1/4) dt

∫ arctan(4x) dx = (1/4) ∫ arctan(t) dt

dv = dt; u = arctan(t)
v=t ; du = 1/(1+t^2)

∫ u dv = u v - ∫ v du
∫ arctan(t) dt = (1/4) t arctan(t) - (1/4) ∫ t dt /(1+t^2)
To integrate ∫ t dt /(1+t^2) , let 1+t^2 = z
2t dt = dz
t dt = (1/2) dz

∫ t dt /(1+t^2) = (1/2) ∫ dz/z = (1/2) ln(z) = (1/2) ln (1+t^2)
∫ arctan(t) dt = (1/4) t arctan(t) - (1/4)(1/2)ln (1+t^2)
最后把t再用x代换回来
=(1/4) 4x arctan(4x) - (1/8) ln (1+ (4x)^2 ) + C
= x arctan(4x) - (1/8) ln (1+ 16x^2 ) + C

[integral] arctan x dx = x arctan x - (1/2) ln(1+x2) + C
x = 4t, dx = 4dt