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

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

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

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

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

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

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

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