Berapa integral dari (x-1)^2 dx

integral (x-1)²
= integral (x-1) (x-1)
= integral (x² - 2x + 1)
= integral 1/2+1 x²⁺¹ - 2/1+1 x¹⁺¹ + x
= 1/3 x³ - x² + x + c

Ada 2 bentuk jawaban tergantung di option nya
1) pakai integral biasa
= Int (x - 1)^2 dx
= Int (x - 1)(x - 1) dx
= Int (x^2 - 2x + 1) dx
= 1/3 x^3 - 2/2 x^2 + x + C
= 1/3 x^3 - x^2 + x + C
2) pakai integral substitusi
= Int (x - 1)^2 dx
= 1/3 (x - 1)^3 + C