Evaluate the integrals in Exercises 31–78.
61. ∫(from 1 ...

Question

Evaluate the integrals in Exercises 31–78.

61. ∫(from 1 to 3)(ln(v+1))²/(v+1) dv

Accepted Answer

Welcome back everyone. Evaluate the integral. Integral from 0 to e of ln 2 of 2 x + 1 divided by 2 x + 1 dx. For this problem we're going to use substitution. What we're going to do is simply assume that u is equal to ln of 2 x + 1. Let's remember that our next step is to differentiate you with respect to x. So we have to apply the chain rule, right? We're going to get 1 divided by 2 x plus 1, multiplied by 2 x plus 1 derivative, and the latter is equal to 2. So we can basically write 2 divided by 2 x plus 1. Now, what we can do is simply multiply both sides by d x and divide both sides by 2. So basically, we, we will get 1 half DU. equals d x. Divided by 2 x plus 1, and this is exactly what we have within the integrant. So now let's change the limits of integration. U1, which is the lower limit, is going to be ln of 2 multiplied by 0 plus 1. That's Ln of 1 which is 0. And now the upper limit u2 is LN of 2 multiplied by e plus 1, or basically 2e plus 1. So now the integral becomes Integral from 0 up to lm of 2e plus 1. Now we have u 2. And we also have the u, right? Essentially ln of 2 x + 1 is u and it is, it is squared. And then we have the x divided by 2 x plus 1, which is 1/2 the u. So all that we have to do is simply add one half in front of the integral. And now we can integrate using the power rule. So that would be 1/2 multiplied by u cubed divided by 3. And we went to evaluate the result from 0 up to LN of 2 E + 1. Let's go ahead and simplify. This is 1/6. You cubed. Evaluated from 0 up to ln of 2e + 1. Now, applying the fundamental theorem of calculus part 2, we are going to get 1/6. In parentheses lm cubed of 2 e + 1 minus. 0 cubed. Now we can exclude 0, and this will essentially give us the final answer, which is 1/6 lm cubed of 2 E plus 1. Thank you for watching.

Evaluate the integrals in Exercises 31–78.61. ∫(from 1 to 3)(ln(v+1))²/(v+1) dv

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Evaluate the integrals in Exercises 31–78.
61. ∫(from 1 to 3)(ln(v+1))²/(v+1) dv