解答
∫(x2+x+1)2x2+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)2x2+2dx
乘开 (x2+x+1)2x2+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