Evaluate the indefinite integral. (remember to use absolute values where appropriate. use c for the constant of integration.) dx hx + q (h ≠ 0

Answer:[tex]\displaystyle \int {(hx + q)} \, dx = \frac{hx^2}{2} + qx + 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:*Note:Assume h and q are arbitrary constants, where h and q ≠ 0.Step 1: DefineIdentify[tex]\displaystyle \int {(hx + q)} \, dx[/tex]Step 2: Integrate[Integral] Rewrite [Integration Property - Addition/Subtraction]:               [tex]\displaystyle \int {(hx + q)} \, dx = \int {hx} \, dx + \int {q} \, dx[/tex][Integrals] Rewrite [Integration Property - Multiplied Constant]:               [tex]\displaystyle \int {(hx + q)} \, dx = h\int {x} \, dx + q\int {} \, dx[/tex][Integrals] Reverse Power Rule:                                                                   [tex]\displaystyle \int {(hx + q)} \, dx = h \Big( \frac{x^2}{2} \Big) + qx + C[/tex]Simplify:                                                                                                         [tex]\displaystyle \int {(hx + q)} \, dx = \frac{hx^2}{2} + qx + C[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)Unit:  Integration