Evaluate the integrals in Exercises 97–110.
107. ∫₀⁹ (2 ...

Question

Evaluate the integrals in Exercises 97–110.

107. ∫₀⁹ (2 log₁₀(x + 1) / (x + 1)) dx

Accepted Answer

Welcome back everyone. Evaluate the integral. Integral from 0 to 99 of 3 log of base 10 of x + 1 divided by x + 1. Aces 2 LN 10, B 6 LN 10, C 4 LN 210, and D 12 ln 210. For this problem to simplify the integral, we're going to perform the change of base of the log. We have log base 10 of X + 1. Let's use the formula and let's change the base toe. So we're going to have log of base e of x + 1, and remember that we're dividing by log of base E of 10, right? And because log of base e can be written as ln, we are going to get ln of x + 1 divided by ln of 10 using the shorthand notation. So now we can rewrite our integral. It is going to be the integral from 0 up to 99 of 3 multiplied by LN of X + 1 divided by Ln of 10, which is multiplied by x + 1. What we can do is factor out the constant, so that's 3 divided by Alan of 10. And we're integrating from 0 to 99. L n of x + 1 divided by x + 1. Now let's introduce a substitution. Let's suppose that u is equal to ln of x + 1. Then the derivative of u with respect to x according to the chain rule is going to be 1 divided by x plus 1 multiplied by the derivative of x plus 1, which is 1, and we can now show that d x. Divided by x plus 1. is simply equal to DU, right? Because we have DX divided by x plus 1 within the integral. This is what we want to rearrange for. And then we're going to change the limits of integration in terms of you. So the lower one is going to be LN of 0 + 1, that's LN of 1, and it is equal to 0. And then we are going to get our upper limit of integration U2 as LN of. 99 + 1, that's LN of 100 or basically 10 squared, right? We're going to rewrite it as Llan of 10 squared, we can apply the power rule to get to LN of 10. And then we can rewrite our integral as integral from 0 up to 2 of 10. We also have our constant 3 divided by ln of 10. So let's write that. In front of the integral. And now in terms of you, we're going to have you. Multiplied by d u. Let's go ahead and integrate. So first of all, we're going to write our constant 3 divided by een of 10 multiplied by the integral of u d u is u 2 divided by 2, so we can write u square and we can add 2 in the denominator of our constant. And now we're going to evaluate from 0 up to 21 of 10. So let's go ahead and substitute. We are going to get 3 divided by 2 ln of 10. And now applying the evaluation theorem, we want to take 2 LN of 10. Squared -02, so there is no necessity to subtract 0, right? We can just eliminate that. And we can now square 2 of 10, which is 4 e square of 10. So we get 3 divided by 2 ln of 10 multiplied by. 4 ln 2 of 10 Or basically we can just cancel out the square, right? Because we have the same term. In the denominator, so let's go ahead and do that. And it allows us to get the answer immediately. Now let's rewrite the final answer. It would be 3 multiplied by 2 of 10, which is. 6 ln of 10 and this corresponds to the answer choice b. Thank you for watching.

Evaluate the integrals in Exercises 97–110.107. ∫₀⁹ (2 log₁₀(x + 1) / (x + 1)) dx

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Evaluate the integrals in Exercises 97–110.
107. ∫₀⁹ (2 log₁₀(x + 1) / (x + 1)) dx