We calculate the integral $\int_{-\infty}^1 (2^x+e^{2x})dx$.
From the table with antiderivatives of elementary functions it follows that $2^x/\ln 2$ is an antiderivative of $2^x$ and that $\frac{1}{2}e^{2x}$ is an antiderivative of $e^{2x}$. Applying the sum rule gives that $F(x)=2^x/\ln 2+\frac{1}{2}e^{2x}$ is an antiderivative of $f(x)=2^x+e^{2x}$.
In order to solve the improper integral $\int_{-\infty}^1 (2^x+e^{2x})dx$, we replace the infinite integration bound by a variable one $t$ and solve the resulting integral.
$$\int_t^1 (2^x+e^{2x})dx=[2^x/\ln 2+\tfrac{1}{2}e^{2x}]_{x=t}^{x=1}=(2/\ln 2+\tfrac{1}{2}e^2)-(2^t/\ln 2+\tfrac{1}{2}e^{2t}).$$
Since $2^t\rightarrow 0$ and $e^{2t}\rightarrow 0$ if $t\rightarrow-\infty$, it follows that
$$(2/\ln 2+\tfrac{1}{2}e^2)-(2^t/\ln 2+\tfrac{1}{2}e^{2t})\rightarrow (2/\ln 2+\tfrac{1}{2}e^2)-(0+0).$$
Conclusion: $\int_{-\infty}^1 (2^x+e^{2x})dx=2/\ln 2+\tfrac{1}{2}e^2$.
