积分∫arctanx*x^2/(1+x^2)dx

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积分∫arctanx*x^2/(1+x^2)dx
应该用分部积分法,可我怎么算也不和答案上一样

∫[arctan(x)]*x^2/(1+x^2)dx = ∫1*arctanxdx-∫(arctanx)/(x^2+1)dx = {x*arctan(x)-∫x/(x^2+1)dx}-∫[arctan(x)]d[arctan(x)] = x*arctan(x)-{ln(x^2+1)+[arctan(x)]^2}/2+C .

求积分 ∫(arctanx)/(x^2(x^2+1))dx 积分∫arctanx*x^2/(1+x^2)dx 定积分∫(-1,1)arctanx/(1+x^2)dx, 定积分∫(x^2arctanx+cos^5x)dx 求定积分[0,1]arctanx/(1+x^2)dx 求定积分∫x^2*(arctanx)^2/(1+x^2)dx (-1 ∫(arctanx)^2/1+X^2 dx ∫(arctanx)^3/(1+x^2)dx ∫(0 1)x(arctanx)^2dx 求定积分∫(-1,1)arctanx/(1+x)^2dx 定积分∫(arctanx)/1+X^2 dx 上限是1,下限是0, 广义积分∫ (正无穷,1) [arctanx/(1+x^2)^3]dx