How do you evaluate the integral #int(1+x)^2 dx#?
1 Answer
Aug 15, 2014
#=x^3/3+x^2+x+C# , where#C# is a constantExplanation :
#I=int(1+x)^2dx# It can be solved by two methods,
#(I)# using Integration by Substitution
let's
#1+x=t# #=># #dx=dt# then,
#intt^2dt=t^3/3+c# Substituting
#t# back,
#=(1+x)^3/3+c# , where#c# is a constant
#=1/3(x^3+3x^2+3x+1)+c# , where#c# is a constant
#=x^3/3+x^2+x+1/3+c# , where#c# is a constant
#=x^3/3+x^2+x+C# , where#C# is again a constant
#(II)# Expanding
#(1+x)^2=x^2+2x+1# , we get
#int(x^2+2x+1)dx#
#=x^3/3+2x^2/2+x+c# , where#c# is a constant
#=x^3/3+x^2+x+c# , where#c# is a constant
