\( \int _ { 0 } ^ { 3 } ( e ^ { 6 } + \operatorname { In } x ) d x \)

To solve the integral \( \int _ { 0 } ^ { 3 } ( e ^ { 6 } + \operatorname { In } x ) d x \), we can break it into two separate integrals: \[ \int _ { 0 } ^ { 3 } e ^ { 6} \, dx + \int _ { 0 } ^ { 3 } \operatorname{ln} x \, dx \] The first integral is straightforward since \( e^6 \) is a constant: \[ \int _ { 0 } ^ { 3 } e ^ { 6} \, dx = e^6 (3 - 0) = 3 e^6 \] The second integral, \( \int _ { 0 } ^ { 3 } \operatorname{ln} x \, dx \), can be solved using integration by parts. Let \( u = \operatorname{ln} x \) and \( dv = dx \). Then, we have \( du = \frac{1}{x}dx \) and \( v = x \). Applying integration by parts gives: \[ \int \operatorname{ln} x \, dx = x \operatorname{ln} x - \int \frac{x}{x}dx = x \operatorname{ln} x - x + C \] Evaluating this from 0 to 3, we have: \[ \left[ x \operatorname{ln} x - x \right]_{0}^{3} \] Calculating it: When \( x = 3 \): \[ 3 \operatorname{ln} 3 - 3 \] When \( x \to 0 \), \( \operatorname{ln} 0 \) approaches \(-\infty\), so we consider the limit: \[ \lim_{x \to 0^+} \left( x \operatorname{ln} x - x \right) = \lim_{x \to 0^+} (x \operatorname{ln} x) - \lim_{x \to 0^+} x = 0 \] Therefore, \[ \left[ x \operatorname{ln} x - x \right]_{0}^{3} = (3 \operatorname{ln} 3 - 3) - 0 = 3 \operatorname{ln} 3 - 3 \] Now, combine both integrals: \[ 3 e^6 + (3 \operatorname{ln} 3 - 3) = 3 e^6 + 3 \operatorname{ln} 3 - 3 \] Thus, the final answer for the integral is: \[ \int _ { 0 } ^ { 3 } ( e ^ { 6 } + \operatorname { In } x ) d x = 3 e^6 + 3 \operatorname{ln} 3 - 3 \]