23-64. Integration Evaluate the following integrals.
62....

Question

23-64. Integration Evaluate the following integrals.

62. ∫ 1/[(x + 1)(x² + 2x + 2)²] dx

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the integral, the interval of 1 divided by parentheses T + 3 multiplied by parentheses T 2 + 6 T + 10 2 DT. Awesome. So it appears for this particular prompt we're asked to evaluate this specific integral and it's given to us. So now that we know that we're evaluating this specific integral. Our first step that we need to take is we need to complete the square in the quadratic. So we need to take T 2 + 6 T + 10, which will be equal to T + 3 to the power of 2 in parentheses, so it's going to be parentheses T + 3 to the power of 2 + 1. So, our next part of this step is we need to determine that the denominator will be parentheses T + 3 in parentheses multiplied by parent of T + 3 for the power of 2 + 1. All to the power of 2. Fantastic. So, moving right along here, our second step now is we need to recall and apply the substitution method. So let us state that you will be equal to T + 3 and DU will be equal to DT. And with that in mind, our integral will become the integral of 1 divided by U, multiplied by parentheses U 2 + 12 D U. fantastic. Our third step now is we need to perform our partial fraction setup. So as we should recall. 41 divided by U multiplied by parenthesis U 2 + 1 to 2, the decomposition will be equal to a divided by U plus BU plus C divided by U 2 + 1 plus DU plus E. Divided by parentheses U 2 + 1 in parenthesis to the power at 2. Fantastic. Our fourth step now is we need to multiply through by U multiplied by parentheses U 2 + 1 in parenthesis to the power of 2, and when we do, we will find that 1 is equal to a multiplied by parentheses U 2 + 12 plus B in parenthesis BU. Plus C multiplied by U, multiplied by parentheses U 2 + 1 + par DU plus E multiplied by U. Fantastic. So our 5th step now is we need to expand the terms starting with our first term. So we're expanding the terms starting with our first term. So the first term states that a multiplied by parentheses U 2 + 1 to 2 in parentheses is equal to a multiplied by parenthesis U 4 + 22 + 1. Awesome. And our second term, We can go ahead and write that parentheses BU plus C multiplied by U, multiplied by parentheses U 2 + 1 will be equal to BU 4 plus BU 2 + Cu 3 + Cu. And for our third term, we can go ahead and write that parentheses DU plus E multiplied by U will be equal to drum roll, DU 2 plus EU. Fantastic. Moving right along here. Our 6th step, so step number 6, is we need to combine the coefficients by powers of U. So U to the power of 4, we could say A plus B, U to the power of 3, we could say C. And moving right along here, we are on a roll. We can say that U to the power of 2 is 2A or 2 multiplied by A, if you want to be specific, but we could say 2A. Plus B plus D. and finally, for you to the power of 1, we can say C plus E. and for our constant value, which I'll just denote as C, we can say A. So, moving right along here. We must have that. U to the power of 4, which is A plus B, must be equal to 0, and we could say that U to the power of 3, C will have to be equal to 0, and u to the power of 2, we could say that 2A plus B plus D will be equal to 0, and for you to the power of 1 C plus E must equal 0, and for Our constant C, which could also be U to the power of 0, A must equal 1. So with that in mind, our seventh step is we need to solve the system. So from where A is equal to 1, we can state that A plus B is equal to 0, so we could say that 1 plus B is equal to 0, and we could say that B is equal to -1, C will be equal to 0. And C + E will be equal to 0, so you could say that E is equal to 0. And 2 A plus B. Plus D, so once again 2A plus B plus D is equal to 0, so we could say 2 multiplied by 1 because we're plugging in a value of 1 in for A. So 2 multiplied by 1 plus 1 + D is equal to 0. So we could say that 1 plus D is equal to 0, so we could say simply that D is equal to -1. Fantastic. So we can safely say that A is equal to 1, B is equal to -1. C is equal to 0, D is equal to -1, and E will be equal to 0. Fantastic. So, moving right along here, step number 8, we need to we need to determine the partial fraction result. So we need to take 1 divided by U multiplied by parentheses U 2 plus 1 in parenthesis to 2, which will be equal to 1 divided by U minus u divided by U 2 plus 1. Minus U divided by parenthesis U2 + 1 to 2. Fantastic. Step number 9 is we need to integrate term by term. So when we integrate term by term, you will find that the integral of 1 divided by U DU will be equal to the natural log of the absolute value of you. We also will find that the integral of negative U divided by U 2 + 1, DU will be equal to -12. Multiplied by the natural log of u to the power 2 + 1 fantastic, and we will determine that the integral of negative u divided by parentheses U 2 + 12 DU. Awesome. So, once again, Moving right along here. Let us state that we'll say that W is equal to U 2 + 1 and that DW is equal to 2UDU. And when we do that, we will find that it will be equal to -12, so negative 1/2. Multiplied by the integral of W to the power of -2 DW, which is equal to 1/2 multiplied by W to the power of -1, which will be equal to 1 divided by 2 multiplied by parenthesis U2 to the power of 1. So U 2 + 1. Multiplied by 2 in the denominator, with 1 in the numerator. Fantastic. So, moving right along here. Step number 10 is we need to determine the final answer in you. So we take the integral of 1 divided by U multiplied by parenthesis U2 plus 1 tou, which is equal to the natural log of the absolute value of you minus 12 multiplied by the natural log of u2 + 1 + 1 divided by 2 multiplied by parenthesis U 2 + 1. Plus C. So now, for our final step for step 11, we need to substitute back in our values. So let's make a note, we're substituting in back our value of U is equal to T + 3. So, when we do that, we will find that the natural log, so we're gonna take the natural log of the absolute value of T + 3 minus 12 multiplied by the natural log of T + 3 to 2 + 1. Plus 1 divided by 2 multiplied by parentheses of T + 3, the power 2 + 1. Plus C will be all equal to, so this will be all equal to when we simplify. So now when we simplify, you will find that it will be equal to the natural log of the absolute value of T + 3 minus 12 multiplied by the natural log of T 2 + 6 T + 10. Plus 1 divided by 2 multiplied by parenthesis T 2 plus 6 T plus 10. Plus C. Fantastic, we did it. We found our final answer. Hooray, we did it. So this big long expression is our final answer for our evaluated interval. So going back to look at the prom itself to recap what we've learned, our final answer, once again looking at the problem itself, our final answer, without a doubt, will have to be the following. So recapping. Our final answer, the natural log of the absolute value of T + 3 minus 12 multiplied by the natural log of T 2 + 6 T + 10. Plus 1 divided by 2 multiplied by parentheses T 2 + 6 T + 10 + C, noting that C represents some constant value, and that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

23-64. Integration Evaluate the following integrals.
62. ∫ 1/[(x + 1)(x² + 2x + 2)²] dx

Key Concepts

Partial Fraction Decomposition

Integration of Rational Functions with Repeated Quadratic Factors

Substitution Method in Integration

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