∫dx/((x^2+1)(x^2+x)

显示全部楼层 · 2020-12-16 18:33:17

∫dx/((x^2+1)(x^2+x)的不定积分是ln│x│-(1/2)ln│x+1│-(1/4)ln(x2+1)-(1/2)arctanx+C。解：∫dx/((x^2+1)(x^2+x)dx=∫[1/x-(1/2)/(x+1)-(x/2)/(x2+1)-(1/2)/(x2+1)]dx=ln│x│-(1/2)ln│x+1│-(1/4)ln(x2+1)-(1/2)arctanx+C所以∫dx/((x^2+1)(x^2+x)的不定积分是ln│x│-(1/2)ln│x+1│-(1/4)ln(x2+1)-(1/2)arctanx+C。扩展资料：1、分部积分法的形式（1）利用有些函数经一次或二次求...

千问 · 2020-12-16 18:33:17

设1/(x^2+1)(x^2+x）=（Ax+B)/(x^2+1)+C/(x+1)+D/x,用待定系数法解之，得：A=-1/2,B=-1/2,C=-1/2,D=1,原式=（-1/2）∫ （x+1)dx/(x^2+1)-(1/2)∫dx/(x+1)+∫dx/x=(-1/4)∫ d(x^2+1)/(x^2+1)-(1/2)∫ dx/(x^2+1...

千问 · 2020-12-16 18:33:17

回答

千问 · 2020-12-16 18:33:17

∫dx/((x^2+1)(x^2+x)dx=∫[1/x-(1/2)/(x+1)-(x/2)/(x2+1)-(1/2)/(x2+1)]dx =ln│x│-(1/2)ln│x+1│-(1/4)ln(x2+1)-(1/2)arctanx+C...