∫4/(1-2x)^2 dx ∫1/(3x+5)dx 利用换元积分法求不定积分~

∫（x^2+1/x^4)dx ∫1/(x^4-x^2)dx ∫x^3/1+x^2 dx ∫(x-1)^2/x^3 dx ∫(X^3)/(1+X^2)dx ∫(x^3-x^2+x+1)/(x^2+1) dx∫(x+4)/(x^2-x-2) dx x-9/[(根号)x]+3 dx ∫ x+1/[(根号)x] dx ∫ [（3-x^2）]^2 dx ∫(3x^4+x^2)/(x^2+1)dx ∫x/(3+4x+x^2)^1/2dx ∫[dx/(e^x(1+e^2x)]dx ∫(x-1)^2dx, ∫x^1/2dx ∫x-根号下x dx ∫lx-2l dx ∫1/根号下（4-x^2) dx ∫e^(-x) dx ∫2/根号下x dx ∫(1/x^2)sin(1/x) dx ∫x^3/(x^8-2) dx∫(x^3-1)/(x^2+1) dx ∫cos x / ( 1 + (sinx)^2 ) dx = ∫x^3 / ( 1 + x^4 ) dx = ∫(sec x)^3 * tan x dx = ∫x^2 * e^(-2x) dx = ∫x * cos 2x dx = ∫(cos 2x)^2 dx = ∫(13x - 6) / ( x (x-2)(x+3) )dx = ∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx = 请把过程写出来哈.>< 旷 这个怎恶魔算∫cos x / ( 1 + (sinx)^2 ) dx = ∫x^3 / ( 1 + x^4 ) dx = ∫(sec x)^3 * tan x dx = ∫x^2 * e^(-2x) dx = ∫x * cos 2x dx = ∫(cos 2x)^2 dx = ∫(13x - 6) / ( x (x-2)(x+3) )dx = ∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx = ∫[(3x+1)/(4+x)^(1/2)]dx高等数学, ∫x^11(2+3x^4)^(1/2)dx