Integrate the function f(x)=3^x+2 with respect to x

First note that a=e^ln(a) and ln(a^b)=bln(a)By substituting a=3^x we get a=3^x=e^{ln(3^x)}=e^xln(3), and hence f(x)=e^xln(3)+2∫ f(x)dx=∫ e^xln(3)+2 dx . First we can split this into the sum of two integrals ∫ e^xln(3)dx + ∫ 2 dxRemember that d/dx(e^g(x)) for some function g is equal to g'(x)e^g(x) by the chain rule so ∫ e^xln(3)dx must equal 1/ln(3)e^xln(3) as 1/ln(3)d/dx(xln(3))=1/ln(3)ln(3)=1And ∫ 2 dx is solved by simply raising the power of any x elements by 1 and dividing the coefficient by this raised power. Hence ∫ 2 dx=2/1x⁰⁺¹=2x=2xSo ∫ f(x)dx=(1/ln(3)e^xln(3))+2x+c (remembering the constant) and by the previous substitution 3^x=e^x*ln(3) ∫ f(x)dx=1/ln(3)*3^x+2x+c