Evaluate the integral. \[ \int \frac{7 e^{7 t}}{e^{14 t}+13 e^{7 t}+36} d t \]

To evaluate the integral \[ I = \int \frac{7 e^{7 t}}{e^{14 t}+13 e^{7 t}+36} d t, \] we first notice that the denominator can be rewritten in terms of a substitution. Let \( u = e^{7t} \). Then, \( du = 7 e^{7t} dt \) or \( dt = \frac{du}{7u} \). Substituting these into the integral, we have: \[ I = \int \frac{7u}{u^2 + 13u + 36} \cdot \frac{du}{7u} = \int \frac{1}{u^2 + 13u + 36} du. \] Next, we factor the quadratic in the denominator. We look for two numbers that multiply to \( 36 \) and add up to \( 13 \). These numbers are \( 9 \) and \( 4 \), so we can factor it as: \[ u^2 + 13u + 36 = (u + 9)(u + 4). \] Now our integral becomes: \[ I = \int \frac{1}{(u + 9)(u + 4)} du. \] To solve this, we use partial fraction decomposition. We can write: \[ \frac{1}{(u + 9)(u + 4)} = \frac{A}{u + 9} + \frac{B}{u + 4}. \] Multiplying through by the denominator \( (u + 9)(u + 4) \), we obtain: \[ 1 = A(u + 4) + B(u + 9). \] Expanding and simplifying gives: \[ 1 = (A + B)u + (4A + 9B). \] To find \( A \) and \( B \), we set up the following system of equations: 1. \( A + B = 0 \) 2. \( 4A + 9B = 1 \) From the first equation, \( B = -A \). Substituting into the second equation: \[ 4A + 9(-A) = 1 \implies 4A - 9A = 1 \implies -5A = 1 \implies A = -\frac{1}{5}. \] Then using \( B = -A \): \[ B = \frac{1}{5}. \] Thus, we can express the integral as: \[ I = \int \left( -\frac{1}{5(u + 9)} + \frac{1}{5(u + 4)} \right) du. \] This separates into two integrals: \[ I = -\frac{1}{5} \int \frac{1}{u + 9} du + \frac{1}{5} \int \frac{1}{u + 4} du. \] Calculating these integrals gives: \[ I = -\frac{1}{5} \ln |u + 9| + \frac{1}{5} \ln |u + 4| + C, \] where \( C \) is a constant of integration. We can simplify this expression using properties of logarithms: \[ I = \frac{1}{5} \left( \ln |u + 4| - \ln |u + 9| \right) + C = \frac{1}{5} \ln \left| \frac{u + 4}{u + 9} \right| + C. \] Finally, we revert back to \( u = e^{7t} \): \[ I = \frac{1}{5} \ln \left| \frac{e^{7t} + 4}{e^{7t} + 9} \right| + C. \] Thus, the evaluated integral is: \[ \int \frac{7 e^{7 t}}{e^{14 t}+13 e^{7 t}+36} d t = \frac{1}{5} \ln \left| \frac{e^{7t} + 4}{e^{7t} + 9} \right| + C. \]