#intx(x+1)^3dx#=?

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Answer 1 · 2018-04-14T14:40:50

#therefore " Option A" #

Approached by #u# substitution...

Let

#u = x+1 #

#u-1 = x #

#=> du = dx #

#=> int (u-1)u^3 du #

#=> int u^4 - u^3 du #

#=> 1/5 u^5 - 1/4 u^4 + c #

#=> 1/5 (x+1)^5 - 1/4 (x+1)^4 + c #

#therefore " Option A" #

Answer 2 · 2018-04-14T14:43:36

The answer is #"option (A)"#

Let #u=x+1#, #=>#, #du=dx#

#x=u-1#

Therefore,

#intx(x+1)^3=int(u-1)u^3du#

#=int(u^4-u^3)du#

#=u^5/5-u^4/4#

#=(x+1)^5/5-(x+1)^4/4+C#

#=(x+1)^4/20((4x+4-5))+C#

#=(x+1)^4/20(4x-1)+C#

The answer is #"option (A)"#