Problem

\( \int \frac{(x-1)(x+4)^{2}}{(x-1)} d x \)

Answer

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Answer

\( \frac{1}{3}x^3 + 4x^2 + 16x + C \)

Steps

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 :\( \frac{1}{3}x^3 + 4x^2 + 16x + C \)

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