(3+2∫2)^(-1/2)=-1+∫2

(3+2∫2)^(-1/2)=-1+∫2 ∫1/[(2-3x)(2x+1)] dx= ∫dx/[(1+x^2)^(3/2)] 设T1=∫∫(x+y)^2dxdy T2=∫∫(x+y)^3dxdy 其中D为(x-2)^2+(y-1)^2 ∫ (2x^2+3)/{(x^2+1)^2} dx 求不定积分 ∫（3x^2-2x+2)dx ∫(2x-1)^2 dx 1/2 ∫ln(1+2x)dx^2=? ∫ x/(1+X^2)dx= ∫xcos(1+x^2)dx= ∫(sinx)^(-1/2)dx= ∫1/(sinx)^2 dx = ∫(1-x)^2/x^3 dx