Let ${\displaystyle \int e^x\left\{f\left(x\right)-f^1\left(x\right)\right\}dx=\varphi \left(x\right)}$ . Then ${\displaystyle \int e^{x}f\left(x\right)dx}$ is

Now, $I={\displaystyle \int e^{x}f\left(x\right)dx}$

$=f\left(x\right){\displaystyle \int e^{x}dx-{\displaystyle \int \left(f^1\left(x\right){\displaystyle \int e^{x}dx}\right)dx}}$

$=f\left(x\right)e^x-{\displaystyle \int f^1\left(x\right).e^{x}dx}$

$=f\left(x\right)e^x-\left[{\displaystyle \int e^{x}f\left(x\right)dx-\varphi \left(x\right)}\right]$

$=f\left(x\right)e^x-I+\varphi \left(x\right)$