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The integral from 1 to 2 of x*the square root of (x-1) dx.

Using the substitution method, I said that u = x-1, du = 1 dx, and dx = du. I'm not sure how to plug u back into the equation since I still have that first x. Suggestions are greatly appreciated.

  • Calculus/Substitution -

    Let u^2=x-1

    INT (u^2+1)u 2udu=
    INT (2u^4+2U^2) du

    revise your limits.

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