int ((1)/(x-2)+sin5x-e^-2x)dx

To solve the integral $\int (\frac {1}{x-2}+sin5x-e^{-2x})dx$, we can break it down into three separate integrals:<br /><br />$\int \frac {1}{x-2}dx + \int sin5x dx - \int e^{-2x} dx$<br /><br />1. $\int \frac {1}{x-2}dx$: This integral is a simple logarithmic integral. The integral of $\frac {1}{x-2}$ with respect to $x$ is $\ln |x-2| + C_1$, where $C_1$ is the constant of integration.<br /><br />2. $\int sin5x dx$: This integral is a standard integral. The integral of $sin5x$ with respect to $x$ is $-\frac{1}{5}cos5x + C_2$, where $C_2$ is the constant of integration.<br /><br />3. $\int e^{-2x} dx$: This integral is also a standard integral. The integral of $e^{-2x}$ with respect to $x$ is $-\frac{1}{2}e^{-2x} + C_3$, where $C_3$ is the constant of integration.<br /><br />Combining the results of the three integrals, we get:<br /><br />$\int (\frac {1}{x-2}+sin5x-e^{-2x})dx = \ln |x-2| - \frac{1}{5}cos5x - \frac{1}{2}e^{-2x} + C$<br /><br />where $C = C_1 + C_2 + C_3$ is the constant of integration.