What is the #intx(x+1)^3"d"x#?

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Answer 1 · 2018-06-30T20:04:16

#intx(x+1)^3dx=1/20(x+1)^4(4x-1)+"c"#

To find #intx(x+1)^3dx#, we use integration by parts

Let #u=xrArrdu=dx#

and #dv=(x+1)^3dxrArrv=1/4(x+1)^4#

Thus

#intx(x+1)^3dx=1/4x(x+1)^4-int1/4(x+1)^4=#

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

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

Answer 2 · 2018-06-30T20:13:36

#I=x^5/5+(3x^4)/4+(3x^3)/3+x^2/2+c#

We know that

#color(blue)((1)(a + b)^3 = a^3 + b^3 + 3ab( a + b )#

Using #(1) # we get

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

#=>I=intx[x^3+1^3+3x(1)(x+1)]dx#

#=>I=intx[x^3+1+3x^2+3x]dx#

#=>I=intx[x^3+3x^2+3x+1]dx#

#=>I=int[x^4+3x^3+3x^2+x]dx#

#=>I=x^5/5+(3x^4)/4+(3x^3)/3+x^2/2+c#