Question

How do you evaluate the integral of `int ln(1+x^3)dx`?

Answer 1

It is

`int ln(1+x^3) dx = x ln(x^3+1)-1/2 ln(x^2-x+1)-3 x+ln(x+1)+sqrt(3) tan^(-1)((2 x-1)/sqrt(3))+c`

Explanation:

Using integration by parts we have that

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

Now for the integral

#int [3x^3]/(1+x^3)dx=int (x-2)/(x^2-x+1)dx-int1/(x+1)+int 3dx= int (x-2)/(x^2-x+1)dx-lnabs(x+1)+3x#

Now for the integral

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

Now we have to calculate the integral as follows

#int 1/[x^2-x+1]dx=int 1/[(x-1/2)^2+3/4]dx= 4/3 int 1/[((x-1/2)/(sqrt3/2))^2+1]dx#

Now we set

`[x-1/2]/[sqrt3/2]=tant=>x-1/2=sqrt3/2*(tant)`

Hence

`dx=sqrt3/2*sec^2t*dt`

Thus now the integral becomes

#4/3* int 1/[tan^2t+1]*(sqrt3/2)*sec^2t*dt= 2/{sqrt3]*tan^-1[(2x-1)/3]#

Finally we get that

`int ln(1+x^3) dx = x ln(x^3+1)-1/2 ln(x^2-x+1)-3 x+ln(x+1)+sqrt(3) tan^(-1)((2 x-1)/sqrt(3))+c`