30pts! Brainliest! Calculus 2!Definite integral of (e^(1/x))/(4x^2) * dxPlease show clear WORK and STEPS for award! Thank you so much!

Answer:[tex]\displaystyle \int {\frac{e^\big{\frac{1}{x}}}{4x^2}} \, dx = \frac{-e^\big{\frac{1}{x}}}{4} + C[/tex]General Formulas and Concepts:CalculusDifferentiationDerivativesDerivative NotationBasic Power Rule:f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Derivative Rule [Quotient Rule]:                                                                           [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]IntegrationIntegrals[Indefinite Integrals] Integration Constant CIntegration Property [Multiplied Constant]:                                                         [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]U-SubstitutionStep-by-step explanation:Step 1: DefineIdentify[tex]\displaystyle \int {\frac{e^\big{\frac{1}{x}}}{4x^2}} \, dx[/tex]Step 2: Integrate Pt. 1[Integral] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle \int {\frac{e^\big{\frac{1}{x}}}{4x^2}} \, dx = \frac{1}{4}\int {\frac{e^\big{\frac{1}{x}}}{x^2}} \, dx[/tex]Step 3: Integrate Pt. 2Identify variables for u-substitution.Set u:                                                                                                             [tex]\displaystyle u = \frac{1}{x}[/tex][u] Differentiate [Derivative Rules, Basic Power Rule]:                               [tex]\displaystyle du = \frac{-1}{x^2} \ dx[/tex]Step 4: Integrate Pt. 3[Integral] Rewrite [Integration Property - Multiplied Constant]:                 [tex]\displaystyle \int {\frac{e^\big{\frac{1}{x}}}{4x^2}} \, dx = \frac{-1}{4}\int {\frac{-e^\big{\frac{1}{x}}}{x^2}} \, dx[/tex][Integral] U-Substitution:                                                                               [tex]\displaystyle \int {\frac{e^\big{\frac{1}{x}}}{4x^2}} \, dx = \frac{-1}{4}\int {e^\big{u}} \, du[/tex][Integral] Exponential Integration:                                                               [tex]\displaystyle \int {\frac{e^\big{\frac{1}{x}}}{4x^2}} \, dx = \frac{-1}{4}e^\big{u} + C[/tex]Simplify:                                                                                                         [tex]\displaystyle \int {\frac{e^\big{\frac{1}{x}}}{4x^2}} \, dx = \frac{-e^\big{u}}{4} + C[/tex]Back-Substitute:                                                                                            [tex]\displaystyle \int {\frac{e^\big{\frac{1}{x}}}{4x^2}} \, dx = \frac{-e^\big{\frac{1}{x}}}{4} + C[/tex]Topic: AP Calculus AB/BC (Calculus I/I + II)Unit: Integration