Question

What is `int x^3ln(x+1+x^2) dx`?

Answer 1

`1/4x^4ln(x^2+x+1)+1/8ln(x^2+x+1)+sqrt3/4arctan((2x+1)/sqrt3)-1/8x^4+1/12x^3+1/8x^2-1/2x+C`

Explanation:

`I=intx^3ln(x^2+x+1)dx`

Use integration by parts (IBP). It takes the form `intudv=uv-intvdu`. For this integral, `intx^3ln(x^2+x+1)dx`, let

`{(u=ln(x^2+x+1),=>,du=(2x+1)/(x^2+x+1)dx),(dv=x^3dx,=>,v=1/4x^4):}`

Thus:

`I=1/4x^4ln(x^2+x+1)-1/4int(2x^5+x^4)/(x^2+x+1)`

Letting `J=int(2x^5+x^4)/(x^2+x+1)`, perform this polynomial long division.

`J=int((-x-2)/(x^2+x+1)+2x^3-x^2-x+2)dx`

Integrating the non-fractional terms first, we see that

`J=int(-x-2)/(x^2+x+1)dx+1/2x^4-1/3x^3-1/2x^2+2x`

Now let `K=int(-x-2)/(x^2+x+1)dx`.

`K=-1/2int(2x+1+3)/(x^2+x+1)=-1/2int(2x+1)/(x^2+x+1)-1/2int3/(x^2+x+1)`

The first can be solved using `t=x^2+x+1` so `dt=(2x+1)`.

`K=-1/2intdt/t-3/2int1/(x^2+x+1)dx=-1/2lnabs(t)-3/2int1/(x^2+x+1)dx`

`K=-1/2ln(x^2+x+1)-3/2int1/(x^2+x+1)dx`

Letting `H=int1/(x^2+x+1)dx`, complete the square in the denominator.

`H=int1/((x+1/2)^2+3/4)dx`

Let `x+1/2=sqrt3/2tantheta`. This implies that `(x+1/2)^2=3/4tan^2theta` and `dx=sqrt3/2sec^2thetad theta`.

`H=int1/(3/4tan^2theta+3/4)(sqrt3/2sec^2thetad theta)`

`H=4/3 sqrt3/2int1/(tan^2theta+1)sec^2thetad theta`

Since `tan^2theta+1=sec^2theta`, this becomes

`H=2/sqrt3intd theta=2/sqrt3theta`

From `x+1/2=sqrt3/2tantheta` we see that `theta=arctan((2x+1)/sqrt3)`.

`H=2/sqrt3arctan((2x+1)/sqrt3)`

Now back-substituting into `K=-1/2ln(x^2+x+1)-3/2H`, we get

`K=-1/2ln(x^2+x+1)-sqrt3arctan((2x+1)/sqrt3)`

This fits into `J=K+1/2x^4-1/3x^3-1/2x^2+2x`, that is,

`J=-1/2ln(x^2+x+1)-sqrt3arctan((2x+1)/sqrt3)+1/2x^4-1/3x^3-1/2x^2+2x`

Finally, since `I=1/4x^4ln(x^2+x+1)-1/4J`, this yields

`I=1/4x^4ln(x^2+x+1)+1/8ln(x^2+x+1)+sqrt3/4arctan((2x+1)/sqrt3)-1/8x^4+1/12x^3+1/8x^2-1/2x+C`