review your properties of definite integrals ...
$\displaystyle \int_1^9 9f(x) \, dx + \int_1^9 8 \, dx = 136$
$\displaystyle 9 \int_1^9 f(x) \, dx = 136 - 8 \int_1^9 \, dx$
$\displaystyle \int_1^9 f(x) \, dx = \frac{1}{9} \left(136 - 8 \int_1^9 \, dx \right)$