How do you find the indefinite integral of int root3x/(root3x-1)∫3√x3√x−1?

1 Answer
1s2s2p
Mar 20, 2018

(root3x-1)^3+(9(root3x-1)^2)/2+9(root3x-1)+3ln(abs(root3x-1))+C(3√x−1)3+9(3√x−1)22+9(3√x−1)+3ln(∣∣3√x−1∣∣)+C

Explanation:

We have int root3x/(root3x-1)dx∫3√x3√x−1dx

Substitute u=(root3x-1)u=(3√x−1)
(du)/(dx)=x^(-2/3)/3dudx=x−233

dx=3x^(2/3)dudx=3x23du

int root3x/(root3x-1)(3x^(2/3))du=int(3x)/(root3x-1)du=int(3(u+1)^3)/udu=3int(u^3+3u^2+3u+1)/udu=int3u^2+9u+9+3/udu=u^3+(9u^2)/2+9u+3ln(abs(u))+C

Resubstitute u=root3x-1:
(root3x-1)^3+(9(root3x-1)^2)/2+9(root3x-1)+3ln(abs(root3x-1))+C

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