17. evaluate the indefinite integral. (int\frac{e^{x}-7x}{7}dx=) _ + c

Explanation:

Step1: Split the integral

We know that $\int\frac{e^{x}-7x}{7}dx=\frac{1}{7}\int(e^{x}-7x)dx$. By the sum - rule of integration $\int(f(x)+g(x))dx=\int f(x)dx+\int g(x)dx$, so $\frac{1}{7}\int(e^{x}-7x)dx=\frac{1}{7}(\int e^{x}dx-\int7xdx)$.

Step2: Integrate each term

The integral of $e^{x}$ with respect to $x$ is $e^{x}$, i.e., $\int e^{x}dx = e^{x}$. The integral of $7x$ with respect to $x$ is $7\int xdx$. Using the power - rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n = 1$, $\int xdx=\frac{x^{2}}{2}$, so $7\int xdx=\frac{7x^{2}}{2}$. Then $\frac{1}{7}(\int e^{x}dx-\int7xdx)=\frac{1}{7}(e^{x}-\frac{7x^{2}}{2})$.

Step3: Simplify the result

$\frac{1}{7}(e^{x}-\frac{7x^{2}}{2})=\frac{e^{x}}{7}- \frac{x^{2}}{2}+C$.

Answer:

$\frac{e^{x}}{7}-\frac{x^{2}}{2}+C$