arctan x/(1+x^2) dx 上限是正无穷下限是1

不定积分arctan(1+x^1/2)dx 求微积分arctan(x^1/2)dx ∫arctan(x/2)dx 求积分∫(arctan(1/x)/(1+x^2))dx 求不定积分∫(arctan(1/x)/(1+x^2))dx ∫arctan^2x/(1+x^2)dx ∫Arctan(1+x^2)dx怎么求? ∫arctan√(x^2-1)dx求不定积分 ∫arctan(1+√x)dx 不定积分arctan根号x dx ∫arctan√x dx ∫(arctan√x)/[√x*(1+x)]dx S(arctan√x)/(√x(1+x))dx 求不定积分arctan(x^1/2)/((x^1/2)×(1+x))dx 求不定积分：∫2x/[(1+x^4)arctan(x^2)] dx ∫dx/(√(x)(1+x))=2arctan(√(x))+c ∫arctan[(x-1)/(x+1)]dx如题 (x+2)arctan(x/2)dx积分怎么求?