\( \int \frac{(x-1)(x+4)^{2}}{(x-1)} d x \)
\( \frac{1}{3}x^3 + 4x^2 + 16x + C \)
Step 1 :\( \int \frac{(x-1)(x+4)^{2}}{(x-1)} d x = \int (x+4)^{2} d x \)
Step 2 :\( \int (x^2 + 8x + 16) d x \)
Step 3 :\( \frac{1}{3}x^3 + 4x^2 + 16x + C \)