(x^2+3)arctanx的不定积分

∫ (x²+3)*arctanx dx = ∫ arctanx d(x³/3 + 3x) = (1/3)∫ arctanx d(x³+9x)
= (x/3)(x²+9)*arctanx - (1/3)∫ (x³+9x) d(arctanx)
= (x/3)(x²+9)*arctanx - (1/3)∫ (x³+9x)/(1+x²) dx
= (x/3)(x²+9)*arctanx - (1/3)∫ [x(x²+1)+8x]/(1+x²) dx
= (x/3)(x²+9)*arctanx - (1/3)∫ x dx - (8/3)∫ x/(1+x²) dx
= (x/3)(x²+9)*arctanx - (1/3)(x²/2) - (8/3)(1/2)∫ d(x²+1)/(x²+1)
= (x/3)(x²+9)*arctanx - (x²/6) - (4/3)ln|x²+1| + C

S(x^2+3)arctanxdx
=Sarctanxd(1/3x^3+3x)
=(1/3*x^3+3x)arctanx-S(1/3*x^3+3x)darctanx
=(1/3*x^3+3x)arctanx-S(1/3*x^3+3x)/(1+x^2)dx
=(1/3*x^3+3x)arctanx-S(1/3*x(x^2+1)+8x/3)/(1+x^2)dx
=(1/3*x^3+3x)arctanx-S(1/3*x)dx-4/3*S1/(x^2+1)d(x^2+1)
=(1/3*x^3+3x)arctanx-1/6*x^2-4/3 *ln(x^2+1)+c

>> int('(x^2+3)*atan(x)',x)

ans =

(x^3*arctan(x))/3 - (4*ln(x^2 + 1))/3 + 3*x*arctan(x) - x^2/6
+C

能求出来的
先将(x^2+3)arctanxdx化为arctanxd(1/3x^3+3x)
然后再利用分部积分就行了呀

原式=∫x^2arctanxdx+3∫arctanxdx对两部分分别用分部积分,∫x^2arctanxdx=∫arctanxd(x^3/3)=(x^3/3)arctanx-（1/3）∫x^3d(arctanx)=(x^3/3)arctanx)-(1/3)∫x^3dx/(1+x^2)=(x^3/3)arctanx-(1/3)∫(x^3+x-x)dx/(1+...