How do you find the indefinite integral of #int root3x/(root3x-1)#?

1 Answer
1s2s2p
Mar 20, 2018

#(root3x-1)^3+(9(root3x-1)^2)/2+9(root3x-1)+3ln(abs(root3x-1))+C#

Explanation:

We have #int root3x/(root3x-1)dx#

Substitute #u=(root3x-1)#
#(du)/(dx)=x^(-2/3)/3#

#dx=3x^(2/3)du#

#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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