49–63. {Use of Tech} Integrating with a CAS Use a computer alg...

Question

49–63. {Use of Tech} Integrating with a CAS Use a computer algebra system to evaluate the following integrals. Find both an exact result and an approximate result for each definite integral. Assume a is a positive real number.
61. ∫₀¹ (ln x) ln(1 + x) dx

61. ∫₀¹ (ln x) ln(1 + x) dx

Accepted Answer

Welcome back, everyone. Compute the exact value of the integral from 12 E of LN T multiplied by a square root of 1 plus L of T squared divided by TDT. A 5 divided by 6, B 2 divided by 3, C, 1 divided by 2, and D 1 divided by 3. So for this problem, first of all, let's focus on a one plus. LM of T. Squared, right? This term is under square root. Whenever we take square root of a square, we get the absolute value of the term that is under the radical. So in this case, we can state that this is equal to the absolute value of 1 plus LN of T. However, we know that t belongs to the interval from 1 toe inclusive. So the natural log extends from LN of 1, which is 0 up to LN of E, which is 1. And then when we add 1, we get from 1 to 2, right? So this value is always positive, meaning we can drop the absolute value and we can show that this is 1 plus Ln of T. We have successfully simplified the integral, and it basically becomes the integral from 1 to E of LN of T multiplied by 1 plus LN of T. Divided by T. DT And to solve for this integral, we're going to introduce the U substitution. Let's suppose that U equals L of T. Then DU divided by DT is going to be the derivative of LN of T, which is 1 divided by T. And we can show that the U is equal to DT divided by C if we multiply both sides by DT. This is exactly what we have. We have DT divided by T. We're going to replace it with DU. And the remaining terms would be Alan of T, which is U1 and Alan of T again, which becomes U. So all that we have to do is Change the limits of integration in terms of you. So the lower limit of integration U1 is going to be Llan of 1. Which is 0, and the upper 1 U2 is LN of E, which is a 1. Meaning we can rewrite our integral as integral from 0 to 1. Of you multiplied by 1 plus U D U. And now we are ready to integrate. Before we integrate, we can apply the distributive property and show that we're integrating. U plus U squared DU from 0 to 1. And now applying the power rule we get U squared divided by 2 plus U cubed divided by 3, evaluated from 0 to 1. Substituting 1, we get 1 square divided by 2, which is 1/2 plus. 1 cubed to divided by 3, that's 1/3. And we're subtracting the value of this function at 0, which is 0, right, because each term contains you. So we basically want to add 1/2 and 1/3. The common denominator is 6. And we get 3 + 2 in the numerator, which is 5. So, the final answer to this problem is a 5 divided by 6. Thank you for watching.

49–63. {Use of Tech} Integrating with a CAS Use a computer algebra system to evaluate the following integrals. Find both an exact result and an approximate result for each definite integral. Assume a is a positive real number.
61. ∫₀¹ (ln x) ln(1 + x) dx

Key Concepts

Definite Integrals

Properties of Logarithmic Functions

Use of Computer Algebra Systems (CAS) for Integration

Watch next

49–63. {Use of Tech} Integrating with a CAS Use a computer algebra system to evaluate the following integrals. Find both an exact result and an approximate result for each definite integral. Assume a is a positive real number.61. ∫₀¹ (ln x) ln(1 + x) dx

Key Concepts

Definite Integrals

Properties of Logarithmic Functions

Use of Computer Algebra Systems (CAS) for Integration

Watch next

49–63. {Use of Tech} Integrating with a CAS Use a computer algebra system to evaluate the following integrals. Find both an exact result and an approximate result for each definite integral. Assume a is a positive real number.
61. ∫₀¹ (ln x) ln(1 + x) dx