Evaluate the integral. (Remember to use absolute values where appropriate. Remember the constant of integration.) \[ \int \frac{37 x+5}{(6 x+1)(x-1)} d x \]

To evaluate the integral \[ \int \frac{37x + 5}{(6x + 1)(x - 1)} \, dx, \] we can use partial fraction decomposition. We can express the integrand in the form: \[ \frac{37x + 5}{(6x + 1)(x - 1)} = \frac{A}{6x + 1} + \frac{B}{x - 1}, \] where \(A\) and \(B\) are constants to be determined. Multiplying both sides by the denominator \((6x + 1)(x - 1)\), we get: \[ 37x + 5 = A(x - 1) + B(6x + 1). \] Expanding the right-hand side, we have: \[ 37x + 5 = Ax - A + 6Bx + B. \] This simplifies to: \[ (6B + A)x + (B - A) = 37x + 5. \] Now, we can equate the coefficients of like terms: 1. For \(x\): \(6B + A = 37\) 2. For the constant: \(B - A = 5\) Now, we can solve this system of equations. From the second equation, we can express \(A\) in terms of \(B\): \[ A = B - 5. \] Substituting this into the first equation gives: \[ 6B + (B - 5) = 37, \] \[ 7B - 5 = 37, \] \[ 7B = 42 \implies B = 6. \] Now substitute \(B\) back to find \(A\): \[ A = 6 - 5 = 1. \] Thus, the partial fractions are: \[ \frac{37x + 5}{(6x + 1)(x - 1)} = \frac{1}{6x + 1} + \frac{6}{x - 1}. \] Now we can rewrite the integral: \[ \int \frac{37x + 5}{(6x + 1)(x - 1)} \, dx = \int \left( \frac{1}{6x + 1} + \frac{6}{x - 1} \right) \, dx. \] This splits into two integrals: \[ \int \frac{1}{6x + 1} \, dx + 6 \int \frac{1}{x - 1} \, dx. \] Each integral can be evaluated separately. The first integral can be solved by a simple substitution \(u = 6x + 1\), \(du = 6 \, dx\), so: \[ \int \frac{1}{6x + 1} \, dx = \frac{1}{6} \ln |6x + 1| + C_1. \] The second integral gives: \[ 6 \int \frac{1}{x - 1} \, dx = 6 \ln |x - 1| + C_2. \] Putting everything together, we combine the constants \(C_1\) and \(C_2\) into a single constant \(C\): \[ \int \frac{37x + 5}{(6x + 1)(x - 1)} \, dx = \frac{1}{6} \ln |6x + 1| + 6 \ln |x - 1| + C. \] Finally, we can express the answer compactly: \[ \int \frac{37x + 5}{(6x + 1)(x - 1)} \, dx = \frac{1}{6} \ln |6x + 1| + 6 \ln |x - 1| + C. \]