请问数分高手们，这个求定积分的题
∫[0,1][(arctanx)/(x+1)]dx
=∫[0,π/4][(y sec^2 y)/(tan y+1)]dy 换元x=tan y
=∫[0,π/4][y/(sin ycos y+cos ^2 y)]dy
=∫[0,π/4][y/(sin ycos y+cos ^2 y)]dy
=∫[0,π/4][(2y )/(sin 2y+cos 2y+1)]dy
=(1/2)∫[0,π/2][x/(sin x+cos x+1)]dx 换元2y=x
令I=∫[0,π/2][x/(sin x+cos x+1)]dx ，则
I=∫[0,π/2][(π/...全部
∫[0,1][(arctanx)/(x+1)]dx
=∫[0,π/4][(y sec^2 y)/(tan y+1)]dy 换元x=tan y
=∫[0,π/4][y/(sin ycos y+cos ^2 y)]dy
=∫[0,π/4][y/(sin ycos y+cos ^2 y)]dy
=∫[0,π/4][(2y )/(sin 2y+cos 2y+1)]dy
=(1/2)∫[0,π/2][x/(sin x+cos x+1)]dx 换元2y=x
令I=∫[0,π/2][x/(sin x+cos x+1)]dx ，则
I=∫[0,π/2][(π/2-y)/(sin y+cos y+1)]dy 换元x=π/2-y
=(π/2)∫[0,π/2][1/(sin y+cos y+1)]dy-I
于是I=(π/4)∫[0,π/2][1/(sin y+cos y+1)]dy
=(π/4)ln2
从而∫[0,1][(arctanx)/(x+1)]dx=(π/8)ln2
。
收起
