Question

Integration of `1/(1+x^3)dx`?

Answer 1

`1/3ln|x+1|-1/6ln|x^2-x+1|+sqrt3/3tan^-1((2x-1)/sqrt3)+C`

Explanation:

Begin by factorizing the denominator:
`1+x^3=(x+1)(x^2-x+1)`

Now we can do partial fractions:
`1/(1+x^3)=1/((x+1)(x^2-x+1))=A/(x+1)+(Bx+C)/(x^2-x+1)`

We can find `A` using the cover-up method :
`A=1/((text(////))((-1)^2+1+1))=1/3`

Next we can multiply both sides by the LHS denominator:
`1=1/3(x^2-x+1)+(Bx+C)(x+1)`

`1=1/3x^2-1/3x+1/3+Bx^2+Bx+Cx+C`

`1=(1/3+B)x^2+(B+C-1/3)x+(C+1/3)`

This gives the following equations:
`1/3+B=0 ->B=-1/3`

`C+1/3=1->C=2/3`

This means that we can rewrite our original integral:
`int\ 1/(1+x^3)\ dx=1/3int\ 1/(x+1)-(x-2)/(x^2-x+1)\ dx`

The first integral can be done using an explicit u-substitution, but it's rather clear that the answer is `ln|x+1|`:
`1/3(ln|x+1|-int\ (x-2)/(x^2-x+1)\ dx)`

We can split the remaining integral into two:
`int\ (x-2)/(x^2-x+1)\ dx=1/2int\ (2x-4)/(x^2-x+1)\ dx=`

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

The reason for the trickery with the multiplying and dividing by `2` is to make the left hand denominator easier to use u-substitution on.

I'll call the left integral Integral 1 and the right integral Integral 2

Integral 1
`int\ (2x-1)/(x^2-x+1)\ dx`

Since we already prepared this integral for substitution, all we need to do is substitute `u=x^2-x+1`, and the derivative is `2x-1`, so we divide by that to integrate with respect to `u`:
`int\ cancel(2x-1)/(cancel(2x-1)*u)\ du=int\ 1/u\ du=ln|u|+C=ln|x^2-x+1|+C`

Integral 2
`3int\ 1/(x^2-x+1)\ dx`

We want to get this integral into the form:
`int\ 1/(1+t^2)\ dt=tan^-1(t)+C`

To do this, we need to complete the square for the denominator:
`x^2-x+1=(x-1/2)^2+k`

`x^2-x+1=x^2-x+1/4+k`

`k=3/4`

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

We want to introduce a u-substitution such that:
`(x-1/2)^2=3/4u^2`

`x-1/2=sqrt3/2u`

`x=sqrt3/2u+1/2`

We multiply by the derivative with respect to `u` to integrate with respect to `u`:
`dx/(du)=sqrt(3)/2`

`3*sqrt3/2int\ 1/(3/4u^2+3/4)\ du=3sqrt3/2*1/(3/4)int\ 1/(u^2+1)\ du=`

`=2sqrt3tan^-1(u)+C=2sqrt3tan^-1((2x-1)/sqrt3)+C`

Completing the original integral
Now that we know the answer to Integral 1 and Integral 2, we can plug them back into the original expression to get our final answer:
`1/3(ln|x+1|-1/2ln|x^2-x+1|+sqrt3tan^-1((2x-1)/sqrt3))+C=`

`=1/3ln|x+1|-1/6ln|x^2-x+1|+sqrt3/3tan^-1((2x-1)/sqrt3)+C`

Answer 2

`1/3ln(x+1)-1/6ln(x^2-x+1)+(sqrt3)/3arctan((2x-1)/sqrt3)+C`

Explanation:

`int dx/(x^3+1)`

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

=`1/3int (3dx)/[(x^2-x+1)*(x+1)]`

=`1/3int (x^2-x+1)/[(x^2-x+1)(x+1)]*dx`-`1/3int (x^2-x-2)/[(x^2-x+1)(x+1)]*dx`

=`1/3int dx/(x+1)`-`1/3int ((x+1)(x-2))/[(x^2-x+1)(x+1)]*dx`

=`1/3ln(x+1)+C-1/3 int (x-2)/(x^2-x+1)*dx`

=`1/3ln(x+1)+C-1/6 int (2x-4)/(x^2-x+1)*dx`

=`1/3ln(x+1)+C-1/6 int (2x-1)/(x^2-x+1)*dx`+`1/6 int 3/(x^2-x+1)*dx`

=`1/3ln(x+1)-1/6ln(x^2-x+1)+C`+`1/2 int dx/(x^2-x+1)`

=`1/3ln(x+1)-1/6ln(x^2-x+1)+C`+`int (2dx)/(4x^2-4x+4)`

=`1/3ln(x+1)-1/6ln(x^2-x+1)+C`+`int (2dx)/((2x-1)^2+3)`

=`1/3ln(x+1)-1/6ln(x^2-x+1)+(sqrt3)/3arctan((2x-1)/sqrt3)+C`