∫(x^4-4x^2+5x-15)/(x^2+1)(x-2) dx=?

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/04/28 18:50:42

∫(x^4-4x^2+5x-15)/(x^2+1)(x-2) dx=?
把式子因数化,变成(?)/(x^2+1)+(?)/(x-2)步,带骤,

∵(x^4-4x^2+5x-15)/[(x^2+1)(x-2)]
=[(x^4+x²-5x²-5)+(5x-10)]/[(x²+1)(x-2)]
=[x²(x²+1)-5(x²+1)+5(x-2)]/[(x²+1)(x-2)]
=[(x²+1)(x²-5)+5(x-2)]/[(x²+1)(x-2)]
=(x²-5)/(x-2)+5/(x²+1)
=[(x²-2x)+(2x-4)-1]/(x-2)+5/(x²+1)
=[x(x-2)+2(x-2)-1]/(x-2)+5/(x²+1)
=x+2-1/(x-2)+5/(x²+1)
∴∫(x^4-4x^2+5x-15)/(x^2+1)(x-2)dx
=∫[x+2-1/(x-2)+5/(x²+1)]dx
=x²/2+2x-ln|x-2|+5arctanx+C,(C是积分常数).

∫(x^4-4x^2+5x-15)/(x^2+1)(x-2) dx=? ∫ [(x^3-2x^2+x+1)/(x^4+5x^2+4)]dx ∫（x^2+1/x^4)dx ∫1/(x^4-x^2)dx ∫2^x*3^x/(9^x-4^x) dx ∫（(x+2)/4x(x^2-1)）dx 积分∫x/[(x^2+1)(x^2+4)]dx ∫(3x^4+x^2)/(x^2+1)dx ∫(x-4x^2+4x^3) (1-x^2)^(1/2) dx=-4∫x^2 * (1-x^2)^(1/2)dx,其中x 计算 ∫（x^4-2x^3+x^2+1）/x（x-1)² dx ∫dx/根号(4x-x^2) ∫ x/(1+X^2)dx=