23-64. Integration Evaluate the following integrals.
44....

Question

23-64. Integration Evaluate the following integrals.

44. ∫₁² 2/[t³(t + 1)] dt

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says determine the following integral, and we're given the integral from 1 to 3 divided by 2 of 3 divided by the quantity of X squad, multiplied by the quantity of X minus 2 in quantity DX. So, the first step that we want to do to solve this problem is to take our anagran and break it up into multiple fractions using partial fractions. They recall for partial fractions that 3 divided by the quantity of X squared multiplied by the quantity of X minus 2 in quantity, going to be equal to a divided by X. Plus B divided by X squared. Plus, the divided by the quantity of X minus 2 in quantity. The next step is to multiply both sides of this expression by the denominator on the left hand side. In doing so, we'll get that 3 is equal to a multiplied by X, multiplied by the quantity of X minus 2 in quantity. Plus, B multiplied by the quantity of X minus 2 in quantity. Plus, the multiplied by. So multiplying everything else on the right hand side, that equation, we will get that 3 is equal to AX2d. Minus 2 A X. Plus BX. Minus 2 B plus the X squared. And collecting like powers of axes, we see that 3 is going to be equal to the quantity of A plus D in quantity, multiplied by X2. Plus, the quantity of B minus to A in quantity multiplied by X. minus 2 B. So, equating the coefficients of each power of X on both sides of the equation, we see that A plus C is equal to 0, since we have no quadratic terms on the left-hand side of the equation. For the linear term, we will have that B minus 2A is equal to 0, and for the constant terms, we will have that minus 2 B is equal to 3. Now we can use this last equation to solve for the value of B by dividing both sides of that equation by minus 2. In doing so, we'll see that B is going to be equal to minus 3 divided by 2. We can then use this value of B to find the value of A using our second equation. So substituting this into our second equation, we'll have minus 3 divided by 2, minus 2A is equal to 0. So adding 3 divided by 2 to both sides of the equation, we'll get that minus 2A is equal to 3 divided by 2, and dividing both sides of the equation by minus 2, we get that A going to be equal to minus 3 divided by 4. And now we can use this value of A in our first equation to find the value of D. And so we'll have minus 3 divided by 4, plus C is equal to 0, and adding 3 divided by 4 to both sides of the equation, we see that C is going to be equal to 3 divided by 4. So this means that our original interval, which is the interval, From 1 to 3 divided by 2. Of 3 divided by the quantity of X squad, multiplied by the quantity of X minus 2, and quantity DX is going to be equal to the integral from 1 to 3 divided by 2. Of minus the quantity of 3 divided by 4 in quantity, divided by X. Multiplied by D X, then we'll have plus the interval from +123 divided by 2. Of minus the quantity of 3 divided by 2 in quantity. Divided by X2, multiplied by D X. And then finally we'll have plus the integral from 1 to 3 divided by 2. Of the quantity of 3 divided by 4 in quantity, divided by The quantity of X minus 2 in quantity D X. So now we just need to integrate each one of these intervals. So this is going to be equal to. For the first interval, we'll have minus 3 divided by 4, multiplied by, and here we're going to be integrating 1 divided by X with respect to X, which, if you recall, will give us the natural log of the absolute value of X. This needs to be evaluated from 1 to 3 divided by 2. Then we'll have minus 3 divided by 2, multiplied by, and here we'll be integrating 1 divided by X2 with respect to X, and recall that that will give us minus 1 divided by X, and this needs to be evaluated from 123 divided by 2. And then we'll have plus 3 divided by 4. Multiplied by, and here we're going to be integrating 1 divided by the quantity of X minus 2 in quantity with respect to X, and recall that that's going to give us the natural log of the absolute value of the quantity of X minus 2 in quantity, and this needs to be evaluated from 123 divided by 2 as well. So substituting in the limits of integration, we see that this is going to be equal to. Minus 3 divided by 4, multiplied by the quantity of the natural log of the quantity of 3 divided by 2 in quantity. Minus the natural log of 1 in quantity. Plus 3 divided by 2, multiplied by the quantity of 1 divided by the quantity of 3 divided by 2 in quantity, minus 1 divided by 1 in quantity. Plus 3 divided by 4, multiplied by the quantity. Of the natural log of the absolute value of the quantity of 3 divided by 2 minus 2 in quantity. Minus the natural log of the absolute value of the quantity of 1 minus 2 in quantity. So now we just need to simplify this expression. So this is going to be equal to. For the first term, we'll have minus 3 divided by 4, multiplied by the natural log of the quantity of 3 divided by 2 in quantity, then we'll have minus 0 for the next term, since the natural log of 1 is 0, then we'll have plus 1. Minus 3 divided by 2, then plus 3 divided by 4, multiplied by the quantity of the natural log of the quantity of 1 divided by 2. In Quanti Minus 0, since the next term is going to be the natural logo of 1, which is 0, in quantity. So, combining like terms, you see that this is going to be equal to. 3 divided by 4, multiplied by the natural log of the quantity of 1 divided by 3 in quantity. -1 divided by 2. Now we can use the properties of logarithms to rewrite the first term. And that is going to be equal to 1 divided by 4 multiplied by the natural log of the quantity of 1 divided by 27 in quantity minus 1 divided by 2. Now we can factor out a minus 1 divided by 4 out of both terms. In doing so, this is going to be equal to -1 divided by 4, multiplied by the quantity of minus the natural log of the quantity of 1 divided by 27 in quantity. Plus 2 in quantity. And again using the properties of logarithms to rewrite this expression, this is going to be equal to -1 divided by 4, multiplied by the quantity of the natural log of 27 plus 2 in quantity. So this here is the answer that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

23-64. Integration Evaluate the following integrals.44. ∫₁² 2/[t³(t + 1)] dt

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23-64. Integration Evaluate the following integrals.
44. ∫₁² 2/[t³(t + 1)] dt