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Answer 1 · 2017-12-20T15:06:33

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

=#(arctanx)^3/3+C#

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

After using #u=arctanx#, #x=tanu# and #dx=(secu)^2*du# transforms, this integral became,

#int (u^2*(secu)^2*du)/(secu)^2#

=#u^2*du#

=#u^3/3+C#

=#(arctanx)^3/3+C#

Answer 2 · 2017-12-20T15:38:51

#1)# #int (x^2-3x)*(cosx)^2*dx#

=#x^3/6-(3x^2)/4+(2x^2-6x-1)/8*sin2x+(2x-3)/8*cos2x+C#

#1)# #int (x^2-3x)*(cosx)^2*dx#

=#1/2int (x^2-3x)*(1+cos2x)*dx#

=#1/2int (x^2-3x)*dx#+#1/2int (x^2-3x)*cos2x*dx#

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

#B=int (x^2-3x)*cos2x#

=#(x^2-3x)*1/2sin2x-(2x-3)(-1/4cos2x)+2(-1/8sin2x)#

=#(2x^2-6x-1)/4*sin2x+(2x-3)/4*cos2x#

Thus,

#int (x^2-3x)*(cosx)^2*dx#

=#1/2*A+1/2*B#

=#x^3/6-(3x^2)/4+(2x^2-6x-1)/8*sin2x+(2x-3)/8*cos2x+C#

Answer 3 · 2017-12-20T15:46:43

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

=#x^2/2+x+(2sqrt3)/3arctan((2x-1)/sqrt3)+C#

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

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

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

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

=#x^2/2+x+(2sqrt3)/3arctan((2x-1)/sqrt3)+C#