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

Answer 1

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

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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Answer 2

Emily Ammerman

To find the indefinite integral of ∫√(3x)/(√(3x) - 1), you can use the substitution method. Let u = √(3x) - 1, then du = (1/2√(3x))dx. Rearrange to solve for dx: dx = 2√(3x)du. Substitute these expressions into the integral and simplify. This will result in a simpler integral in terms of u. Integrate with respect to u, then replace u with the original expression involving x. Finally, add the constant of integration, usually denoted as C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.