∫6,4 （2x+40）dx=（x^2+40x）6,4 这是如何化简的啊

∫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 = ∫（(x+2)/4x(x^2-1)）dx ∫ x（cosx）^2 dx=? ∫lnx/（x^2）dx=? ∫dx/x(x^6+4)=? ∫(√(x^2+6x))dx ∫x^6/1+x^2dx ∫（x^2+1/x^4)dx ∫dx/根号(4x-x^2) ∫1/(x^4-x^2)dx ∫x/(4+9x^2)dx ∫ x/(1+X^2)dx= ∫(x+1/x)^2dx=? ∫（x-3x+2）dx ∫x/（x^2+5）dx 求∫x-3/x²-2x+2 dx,∫x³/√（4-x²）dx ∫[(x^2-x+6)/(x^3+3x)]dx