Question

Determine whether the integral is convergent or divergent. 1 35 ln x x dx 0 convergent divergent Incorrect: Your answer is incorrect.

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The integral  is divergent

Step-by-step explanation:

From the question we are told that

   The equation given is  

                   $\int\limits^{\infty}_1 {35 \frac{ln(x)}{x} } \, dx$

Let    $v = ln x$

=>    $\frac{dv}{dx} = \frac{1}{x}$

=>    $du = \frac{dx}{x}$

So  

         $\int\limits^{\infty}_1 {35 \frac{ln(x)}{x} } \, dx = 35 \int\limits^{\infty}_1 { u} \, du$

=>     $\int\limits^{\infty}_1 {35 \frac{ln(x)}{x} } \, dx = 35 [\frac{u^2}{2} ] | \left {\infty } } \atop {1}} \right.$

=>     $\int\limits^{\infty}_1 {35 \frac{ln(x)}{x} } \, dx = \frac{35}{2} [(ln (x))^2] | \left {\infty } } \atop {1}} \right.$

=>     $\int\limits^{\infty}_1 {35 \frac{ln(x)}{x} } \, dx = \frac{35}{2} [ [(ln (\infty))^2] - [(ln (1))^2] ]$

=>     $\int\limits^{\infty}_1 {35 \frac{ln(x)}{x} } \, dx = \frac{35}{2} [ \infty - [(ln (1))^2] ]$

=>     $\int\limits^{\infty}_1 {35 \frac{ln(x)}{x} } \, dx = \frac{35}{2} [ \infty ]$

=>     $\int\limits^{\infty}_1 {35 \frac{ln(x)}{x} } \, dx = \infty$

Hence given that the solution to the integral  is $\infty$  then it mean that the integral is divergent