\( \int _ { 0 } ^ { \ln 2 } \frac { e ^ { - x } } { e ^ { - x } + 4 } d x \)

To evaluate the integral \( \int _ { 0 } ^ { \ln 2 } \frac { e ^ { - x } } { e ^ { - x } + 4 } d x \), we can use a substitution to simplify the expression. Let's set \( u = e^{-x} \). Therefore, when \( x = 0 \), \( u = 1 \), and when \( x = \ln 2 \), \( u = \frac{1}{2} \). The differential becomes \( du = -e^{-x} dx \) or \( dx = -\frac{du}{u} \). The integral then transforms to: \[ \int _ { 1 } ^ { \frac{1}{2} } \frac{u}{u + 4} \left( -\frac{du}{u} \right) = \int _ { \frac{1}{2} } ^ { 1 } \frac{1}{u + 4} du \] Now we can integrate \( \frac{1}{u + 4} \): \[ \int \frac{1}{u + 4} du = \ln |u + 4| + C \] Evaluating this from \( \frac{1}{2} \) to \( 1 \): \[ \left[ \ln(1 + 4) - \ln\left(\frac{1}{2} + 4\right) \right] = \ln(5) - \ln\left(\frac{9}{2}\right) = \ln\left(5 \cdot \frac{2}{9}\right) = \ln\left(\frac{10}{9}\right) \] Thus, the value of the integral is: \[ \int _ { 0 } ^ { \ln 2 } \frac { e ^ { - x } } { e ^ { - x } + 4 } d x = \ln\left(\frac{10}{9}\right) \]