91Ó°ÊÓ

The integral looks like: \[ \int_{1}^{\infty} \frac{1}{x(x+1)} \, dx \:. \] Let's perform partial fraction decomposition to simplify the integrand: \[ \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \:. \] We can rewrite the equation as: \[ 1 = A(x+1) + Bx \:. \] By setting x = 0, we get: \[ 1 = A \implies A =1 \:. \] By setting x = -1, we get: \[ 1 = -B \implies B = -1 \:. \] Thus, \[ \frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1} \:. \]

Now, we can rewrite the integral as: \[ \int_{1}^{\infty} \frac{1}{x(x+1)} \, dx = \int_{1}^{\infty} \left(\frac{1}{x} - \frac{1}{x+1}\right) \, dx \:. \] Evaluate the difference of two integrals: \[ \int_{1}^{\infty} \left(\frac{1}{x} - \frac{1}{x+1}\right) \, dx = \int_{1}^{\infty} \frac{1}{x} \, dx - \int_{1}^{\infty} \frac{1}{x+1} \, dx \:. \] Now, let's compute each integral separately. The first one: \[ \int_{1}^{\infty} \frac{1}{x} \, dx = \lim_{b\to\infty} \left[ \ln |x| \right]_1^b = \lim_{b\to\infty} (\ln b - \ln 1) = \infty. \] The second one: \[ \int_{1}^{\infty} \frac{1}{x+1} \, dx = \lim_{b\to\infty} \left[ \ln |x+1| \right]_1^b = \lim_{b\to\infty} (\ln (b+1) - \ln 2) = \infty. \] Since both integrals go to infinity, their difference also diverges: \[ \int_{1}^{\infty} \frac{1}{x(x+1)} \, dx = \infty - \infty = \text{diverges} \:. \] Therefore, the given statement is:

False. The improper integral \(\int_{1}^{\infty} \frac{d x}{x(x+1)}\) is divergent, meaning that it does not have a finite value.