Answer
$$ \displaystyle\int \dfrac{{x}^{3}}{{\left(x-1\right)}^{4}}\, \mathrm d x = {{6\,\ln \left(x-1\right)\,x^3-18\,\ln \left(x-1\right)\,x^2-18\, x^2+18\,\ln \left(x-1\right)\,x+27\,x-6\,\ln \left(x-1\right)-11 }\over{6\,\left(x-1\right)^3}} $$
Explanation
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