∫sin2xcos3xdx＝?∫（1－x）÷√(9－4x)dx＝?∫（arctan√x）÷(√x(1＋x))dx＝?

∫sin2xcos3xdx＝?∫（1－x）÷√(9－4x)dx＝?∫（arctan√x）÷(√x(1＋x))dx＝? ∫(arctan√x)/[√x*(1+x)]dx ∫arctan(1+√x)dx ∫(arctan√x)/√x dx ∫arctan[(x-1)/(x+1)]dx 求不定积分∫（arctan√x）÷（√x（x+1））dx,要详细步骤,谢谢. ∫arctan√(x^2-1)dx求不定积分 ∫1/√x*(4-x)dx ∫dx/[x√(1-x^4)] 求积分∫(arctan(1/x)/(1+x^2))dx 求∫arctan（e^x）/（e^x）dx? ∫arctan^2x/(1+x^2)dx