Question

\( \int \left( \frac{1}{7} - 6x - x^2 \right) \, dx \)

Answer 1

1/8 log{7+x/1-x}+c

Answer 2

To integrate \( \int \left( \frac{1}{7} - 6x - x^2 \right) \, dx \), we will integrate each term separately.
1. Integrate \( \frac{1}{7} \) with respect to \( x \):
  \[ \int \frac{1}{7} \, dx = \frac{1}{7}x \]
2. Integrate \( -6x \) with respect to \( x \):
  \[ \int -6x \, dx = -\frac{6}{2}x^2 = -3x^2 \]
3. Integrate \( -x^2 \) with respect to \( x \):
  \[ \int -x^2 \, dx = -\frac{1}{3}x^3 \]
Now, combine all these results to find the definite integral:
\[ \int \left( \frac{1}{7} - 6x - x^2 \right) \, dx = \frac{1}{7}x - 3x^2 - \frac{1}{3}x^3 + C \]
where \( C \) is the constant of integration.
So, the evaluated integral \( \int \left( \frac{1}{7} - 6x - x^2 \right) \, dx \) is \( \frac{1}{7}x - 3x^2 - \frac{1}{3}x^3 + C \).