解答
∫(x2+x+1)2−x2−2dx
解答
−334arctan(31(2x+1))+331sin(2arctan(31(2x+1)))−2(x2+x+1)1−338arctan(31(2x+1))−334sin(2arctan(31(2x+1)))+C
求解步骤
∫(x2+x+1)2−x2−2dx
乘开 (x2+x+1)2−x2−2:−(x2+x+1)2x2−(x2+x+1)22
使用积分加法定则: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=−∫(x2+x+1)2x2dx−∫(x2+x+1)22dx
∫(x2+x+1)2x2dx=332arctan(31(2x+1))+332arctan(31(2x+1))+331sin(2arctan(31(2x+1)))−332sin(2arctan(31(2x+1)))+2(x2+x+1)1
∫(x2+x+1)22dx=334(2arctan(32x+1)+sin(2arctan(32x+1)))
=−(332arctan(31(2x+1))+332arctan(31(2x+1))+331sin(2arctan(31(2x+1)))−332sin(2arctan(31(2x+1)))+2(x2+x+1)1)−334(2arctan(32x+1)+sin(2arctan(32x+1)))
化简 −(332arctan(31(2x+1))+332arctan(31(2x+1))+331sin(2arctan(31(2x+1)))−332sin(2arctan(31(2x+1)))+2(x2+x+1)1)−334(2arctan(32x+1)+sin(2arctan(32x+1))):−334arctan(31(2x+1))+331sin(2arctan(31(2x+1)))−2(x2+x+1)1−338arctan(31(2x+1))−334sin(2arctan(31(2x+1)))
=−334arctan(31(2x+1))+331sin(2arctan(31(2x+1)))−2(x2+x+1)1−338arctan(31(2x+1))−334sin(2arctan(31(2x+1)))
解答补常数=−334arctan(31(2x+1))+331sin(2arctan(31(2x+1)))−2(x2+x+1)1−338arctan(31(2x+1))−334sin(2arctan(31(2x+1)))+C