Evaluate the integrals in Exercises 111–114.
111. ∫₁^(ln...

Question

Evaluate the integrals in Exercises 111–114.

111. ∫₁^(ln x) (1 / t) dt,x > 1

Accepted Answer

Welcome back, everyone. Evaluate the integral from 1 up to LN of X2 + 1 of 1 divided by TDT where X is not equal to 0. For this problem, let's go ahead and rewrite the integral from 1 up to LN of X2 + 1 of 1 divided by TDT. Let's begin by evaluating the integral. This is an elementary integral, and we can conclude that its value is LN of the absolute value of C. And now we want to apply the fundamental theorem of calculus part 2 to evaluate the result from 1 up to LN of X2 + 1. Applying the evaluation theorem, we get LN of the absolute value of LN of X2 + 1 minus LN of the absolute value of 1. Now, what we can do is simply drop each absolute value because one is greater than 0, right? And for LN of X2 + 1, let's understand that if X is not equal to 0, then X2 is always greater than 0, right? So, essentially, We will have X2 + 1, always greater than 1, right? And therefore, if we take LN of a value that is greater than 1, we will get a positive value. So we can drop the absolute value in each case. And now we can begin simplifying, right? So, LN of 1 is equal to 0, we can cross out this term, and therefore, we can conclude that the correct answer is LN of LN X2 + 1. Thank you for watching.

Evaluate the integrals in Exercises 111–114.111. ∫₁^(ln x) (1 / t) dt,x > 1

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Evaluate the integrals in Exercises 111–114.
111. ∫₁^(ln x) (1 / t) dt,x > 1