91Ó°ÊÓ

To make the problem more manageable, let's perform a substitution: $$v = e^u \Rightarrow d v = e^ud u$$ Now our integral becomes: $$\int_{1}^{\infty} \frac{1}{v(v^2 +1)} dv$$ where the lower bound is now \(1\) since \(e^0 = 1\)

To integrate the rational function, we will use partial fractions. $$\frac{1}{v(v^2 +1)} = \frac{A}{v} + \frac{Bv + C}{v^2 + 1}$$ Multiplying both sides by \(v(v^2 +1)\), we get: $$1 = A(v^2 + 1) + (Bv + C)v$$ Now we will identify the coefficients \(A\), \(B\), and \(C\). Setting \(v=0\), we find \(A\): $$1 = A(0^2 + 1)$$ $$A = 1$$ Now we need to find \(B\) and \(C\). Since we already know that \(A=1\), we may choose any other value for \(v\)- let's use \(v=1\): $$1 = (1)(1^2 + 1) + (B + C)(1)$$ $$1-2 = B + C$$ $$B+C=-1$$ Choose another value for \(v\), let's choose \(v=-1\): $$1 = (1)(1^2 + 1) + (-1)(B - C)$$ $$1-2 = -B+C$$ $$-1 = -B+C$$ Solve the system of linear equations of \(B\) and \(C\): $$B+C=-1$$ $$-B+C=-1$$ Adding these equations, we get \(C=-1\), and then substituting back into one of the equations gives us \(B=0\). So, our partial fraction decomposition is: $$\frac{1}{v(v^2 +1)} = \frac{1}{v} - \frac{1}{v^2 + 1}$$

Now we can integrate both terms separately: $$\int_{1}^{\infty} \left( \frac{1}{v} - \frac{1}{v^2 + 1} \right) dv = \int_{1}^{\infty} \frac{1}{v} dv - \int_{1}^{\infty} \frac{1}{v^2 + 1} dv$$ We can easily integrate the first term: $$\int_{1}^{\infty} \frac{1}{v} dv = \left[\ln{v}\right]_{1}^{\infty}$$ This integral diverges because the natural logarithm approaches infinity as \(v\) goes to infinity, so it doesn't have a finite value. Since the first integral diverges, the whole integral diverges. Therefore, we conclude that the given integral diverges.