find the integral.\n int (x^{2}-7x + 5)mathrm{d}x\n int (x^{2}-7x + 5)mathrm{d}x=square

Explanation:

Step1: Apply sum - difference rule

$\int (x^{2}-7x + 5)dx=\int x^{2}dx-\int7xdx+\int5dx$

Step2: Use power rule for $\int x^{n}dx$

For $\int x^{2}dx$, by the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int x^{2}dx=\frac{x^{3}}{3}+C_1$. For $\int7xdx$, since $\int kf(x)dx=k\int f(x)dx$ ($k$ is a constant), $\int7xdx = 7\int xdx=7\times\frac{x^{2}}{2}+C_2=\frac{7x^{2}}{2}+C_2$. For $\int5dx$, as $\int kdx=kx + C$ ($k$ is a constant), $\int5dx=5x+C_3$.

Step3: Combine the results

$\int (x^{2}-7x + 5)dx=\frac{x^{3}}{3}-\frac{7x^{2}}{2}+5x + C$ (where $C = C_1 - C_2+C_3$)

Answer:

$\frac{x^{3}}{3}-\frac{7x^{2}}{2}+5x + C$