Question

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

Answer 1

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

Explanation:

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

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

Explanation:

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`

Using Power Rule:

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