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

Question

Charles Anctil · Accepted Answer

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# 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#

Emily Ammerman · Answer

undefined To evaluate the integral of $ \int \ln(1+x^3) \, dx $, we can use integration by parts with $ u = \ln(1+x^3) $ and $ dv = dx $. Then, we differentiate $ u $ and integrate $ dv $ to find $ du $ and $ v $. After that, we apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

This gives us the integral:

$$ \int \ln(1+x^3) \, dx = x \ln(1+x^3) - \int x \left( \frac{3x^2}{1+x^3} \right) \, dx $$

We can further simplify the remaining integral by making a substitution, letting $ t = 1 + x^3 $ and $ dt = 3x^2 \, dx $, yielding:

$$ \int x \left( \frac{3x^2}{1+x^3} \right) \, dx = \int \frac{dt}{3} $$

Therefore,

$$ \int \ln(1+x^3) \, dx = x \ln(1+x^3) - \frac{1}{3} \int dt $$

$$ = x \ln(1+x^3) - \frac{t}{3} + C $$

Where $ C $ is the constant of integration. Finally, we substitute back $ t = 1 + x^3 $:

$$ = x \ln(1+x^3) - \frac{1+x^3}{3} + C $$