题目

积分∫dx(2x+3)/x^2+3x-10=?请尽可能详细.

提问时间：2021-01-02

答案

原式=∫dx[(x-2)+(x+5)]/(x-2)(x+5)
=∫dx[1/(x-2)+1/(x+5)]
=∫dx/(x-2)+∫dx/(x+5)
=ln(x-2)+ln(x+5)
=ln(x-2)(x+5)
=ln(x^2+3x-10)