What is #int (-2x^3-x^2+x+2 ) / (3x^2+2x +3 )#?

Question

1337 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-17T18:26:20

#(25Ln(3x^2+2x+3)+10sqrt2arctan((3x+1)/(2sqrt2))-18x^2+6x)/54+C# or

=#25/54Ln(3x^2+2x+3)+(5sqrt2)/27arctan((3x+1)/(2sqrt2))-(3x^2-x)/9+C#

#int (-2x^3-x^2+x+2)/(3x^2+2x+3)*dx#

=#-1/9int (18x^3+9x^2-9x-18)/(3x^2+2x+3)*dx#

=#-1/9int ((6x-1)*(3x^2+2x+3)-(25x+15))/(3x^2+2x+3)*dx#

=#-1/9int (6x-1)dx#+#1/9int (25x+15)/(3x^2+2x+3)*dx#

=#-1/9*(3x^2-x)+C#+#1/9int (25/6*(6x+2)+20/3)/(3x^2+2x+3)*dx#

=#-1/9*(3x^2-x)+C#+#25/54int (6x+2)/(3x^2+2x+3)*dx#+#20/27int (dx)/(3x^2+2x+3)#

=#-1/9*(3x^2-x)+25/54ln(3x^2+2x+3)+C#+#20/27int (3dx)/(9x^2+6x+9)#

=#-1/9*(3x^2-x)+25/54ln(3x^2+2x+3)+C#+#20/27int (3dx)/((3x+1)^2+(2sqrt2)^2)#

=#-1/9*(3x^2-x)+25/54ln(3x^2+2x+3)#+#(5sqrt2)/27arctan((3x+1)/(2sqrt2))+C#

=#(25Ln(3x^2+2x+3)+10sqrt2arctan((3x+1)/(2sqrt2))-18x^2+6x)/54+C#