Problem
\( \int \frac{(x-1)(x+4)^{2}}{(x-1)} d x \)
Solution
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 ：\( \frac{1}{3}x^3 + 4x^2 + 16x + C \)
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 ：\( \frac{1}{3}x^3 + 4x^2 + 16x + C \)
