What is #int (-3x^3+4 ) / (-2x^2+x +1 )#?

Question

1203 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-22T18:20:34

#int ((-3x^3+4)*dx)/(-2x^2+x+1)#

=#(3x^2+3x)/4+35/24*Ln(2x+1)-1/3*Ln(x-1)+C#

#int ((-3x^3+4)*dx)/(-2x^2+x+1)#

=#int ((3x^3-4)*dx)/(2x^2-x-1)#

=#1/4*int ((12x^3-16)*dx)/(2x^2-x-1)#

=#1/4*int ([(2x^2-x-1)*(6x+3)+9x-13]*dx)/(2x^2-x-1)#

=#1/4*int (6x+3)*dx+1/4*int ((9x-13)*dx)/(2x^2-x-1)#

=#(3x^2+3x)/4#+#1/4*int ((9x-13)dx)/[(2x+1)(x-1)]#

=#(3x^2+3x)/4+1/4*[35/3*int dx/(2x+1)-4/3*int dx/(x-1)]#

=#(3x^2+3x)/4+35/12*int dx/(2x+1)-1/3*int dx/(x-1)#

=#(3x^2+3x)/4+35/24*Ln(2x+1)-1/3*Ln(x-1)+C#