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

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Answer 1 · 2016-03-02T17:48:19

It is

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

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

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

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

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