Find the general indefinite integral. (use c for the constant of integration.) (7x2 + 8x−2) dx

Answer:[tex]\displaystyle \int ({7x^2 + 8x - 2)} \, dx = \frac{7x^3}{3} + 4x^2 - 2x + C[/tex]General Formulas and Concepts:CalculusIntegrationIntegrals[Indefinite Integrals] Integration Constant CIntegration Rule [Reverse Power Rule]:                                                               [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]Integration Property [Multiplied Constant]:                                                         [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]Integration Property [Addition/Subtraction]:                                                       [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]Step-by-step explanation:Step 1: DefineIdentify[tex]\displaystyle \int ({7x^2 + 8x - 2)} \, dx[/tex]Step 2: Integrate[Integral] Rewrite [Integration Property - Addition/Subtraction]:               [tex]\displaystyle \int ({7x^2 + 8x - 2)} \, dx = \int {7x^2} \, dx + \int {8x} \, dx - \int {2} \, dx[/tex][Integrals] Rewrite [Integration Property - Multiplied Constant]:               [tex]\displaystyle \int ({7x^2 + 8x - 2)} \, dx = 7\int {x^2} \, dx + 8\int {x} \, dx - 2\int {} \, dx[/tex][Integrals] Reverse Power Rule:                                                                   [tex]\displaystyle \int ({7x^2 + 8x - 2)} \, dx = 7 \Big( \frac{x^3}{3} \Big)+ 8 \Big( \frac{x^2}{2} \Big) - 2x + C[/tex]Simplify:                                                                                                         [tex]\displaystyle \int ({7x^2 + 8x - 2)} \, dx = \frac{7x^3}{3} + 4x^2 - 2x + C[/tex]  Topic: AP Calculus AB/BC (Calculus I/I + II)Unit: Integration