∫x^2/(9+4x^2)dx

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/09/20 05:54:31

∫x^2/(9+4x^2)dx

∫x^2/(9+4x^2)dx =1/4* ∫4x^2/(9+4x^2)dx
=1/4 * ∫[1-9/(9+4x^2)]dx
=1/4 *∫dx -1/4 * ∫[ 1/(1+(2x/3)^2)]dx
=1/4 *∫dx -1/4 * 3/2* ∫[ 1/(1+(2x/3)^2)]d(2x/3)
=x/4 - 3/8 * arctan(2x/3) + c

∫2^x*3^x/(9^x-4^x) dx ∫x^3/9+X^2 dx. ∫x^3/(9+x^2)dx ∫（x^2+1/x^4)dx ∫dx/根号(4x-x^2) ∫1/(x^4-x^2)dx x-9/[(根号)x]+3 dx ∫ x+1/[(根号)x] dx ∫ [（3-x^2）]^2 dx ∫x^2/(9+4x^2)dx ∫(1-x)/[√(9-4x^2)]dx= 求∫ (x-1)/ （9-4x^2）dx ∫(x-1)/√(9-4x^2)dx不定积分 ∫（(x+2)/4x(x^2-1)）dx