Mathematics

Mathematics

Problem

Evaluate ∫(2x+1) e3x dx\displaystyle \int (2x+1)\,e^{3x}\,dx. a) e3x(2x+1)3+C\dfrac{e^{3x}(2x+1)}{3}+C b) e3x(6x−1)9+C\dfrac{e^{3x}(6x-1)}{9}+C c) e3x(6x+1)9+C\dfrac{e^{3x}(6x+1)}{9}+C d) e3x(2x−1)+Ce^{3x}(2x-1)+C
Intuitive explanation

This is integration by parts, where one function gets simpler when you differentiate it and the other stays manageable when you integrate it. Choose u=2x+1u = 2x+1 (which becomes the constant 2 when you differentiate) and dv=e3x dxdv = e^{3x}\,dx (which stays exponential when integrated). The key is that the linear part vanishes after one application, leaving you with a straightforward exponential integral to finish.

Intuitive explanation

This is integration by parts, where one function gets simpler when you differentiate it and the other stays manageable when you integrate it. Choose u=2x+1u = 2x+1 (which becomes the constant 2 when you differentiate) and dv=e3x dxdv = e^{3x}\,dx (which stays exponential when integrated). The key is that the linear part vanishes after one application, leaving you with a straightforward exponential integral to finish.

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