92–98. Evaluate the following integrals.
94. ∫ (dt / (t³...

Question

92–98. Evaluate the following integrals.

94. ∫ (dt / (t³ + 1))

Accepted Answer

Hello. In this video, we are going to be solving the following indefinite integral. So we are given the integral of DX divided by X cubed minus 1. Now, we can simplify this integral by first expanding the denominator. We can expand the denominator if we were to expand it by using a difference of cubed terms. If we were to use the difference of cube terms, we can rewrite the given integral to be the integral of 1 divided by X minus 1 multiplied by X2 + X + 1. Now, DX. Now, we can further break down this integral by using partial fraction decomposition. So, we are going to start by setting up the partial fractions. This is going to equal to the partial fractions of A divided by X minus 1, plus B X + C divided by X2 + X + 1. Now, we will go ahead and do the partial fraction decomposition on the side here. If we were to multiply all three terms by the least common denominator, we will be left with 1 is equal to a multiplied by X2 + X + 1. Plus the quantity B X plus C multiplied by X minus 1. Expanding this further will leave us with AX2 + A X + 1 plus BX2 minus BX plus CX minus C. And grouping up like terms and factoring out the respective X terms is going to leave us with 1 is equal to A plus B multiplied by X2, plus a minus B plus C multiplied by X, plus the constant terms A minus C. Now, what we're going to do here is we're going to set up a system of linear equations. We are going to allow our constant terms to equal to the constants on both the left and right hand side, so that is going to give us a minus C is equal to 1 as our first equation. We are going to set both of the X coefficients equal to each other, and so that's going to be a minus B plus C is equal to 0, since we have no X term on the left hand side. And we're going to set our X2 terms equal to each other, which is going to be A plus B is equal to 0, since we have no X2 terms on the left hand side. Now, if we were to solve the following system, we will get the resulting values for A, B, and C. We will get that A is equal to 1/3, B is equal to -1/3, and C is equal to 2/3. So, if we were to plug this back into the partial fraction decomposition we wrote earlier, we will end up with 1/3 divided by X minus 1. Plus 1/3 multiplied by X minus 2/3 divided by the quantity X2 + X + 1. Now, all we've done so far is just simplify the integrand using algebra, which means that we have to include this underneath the integral symbol. In fact, what we can do is we can further simplify this into a sum of 3 integrals. So if we were to further simplify, we would have the integral of 1 divided by 3 multiplied by X minus 1, minus the integral of 1 divided. Or 1 X, which is going to be X divided by 3, multiplied by X2 + X + 1. Minus the integral of 2 divided by 3 multiplied by X2 + X + 1. OK, so now what we're going to do is we're going to evaluate each of the integrals independently, and then combine the values later. So let's go ahead and start with the first integral. Now, the first integral can be rewritten as 1/3, multiplied by the integral of 1 divided by X minus 1. And by using our natural logarithm property, this is going to evaluate to be 1/3 multiplied by the natural logarithm of X minus 1. This is going to be the value of the first integral, and we will set it to the side for now. Now, for the second integral, we have negative. Integral of X divided by 3 multiplied by X2 + X + 1. Now, if we were to pull the 3 out as a constant, we will end up with negative 1/3 as a constant in front of the integral. Now, here we are going to approach this by using a U substitution. Let's go ahead and allow U to equal to X2 + X + 1. Then DU we equal to 2 X plus 1 DX. Now, the problem here is that we don't have 2 X plus 1 inside of our integral at all in order to make a substitution for DX. So instead, what we're going to need to do is we're going to need to manipulate our numerator in order to get a 2 X + 1 term. Now, if we were to manipulate our numerator, we can rewrite the integral to be 1/2 multiplied by 2 X + 1 minus 1 divided by X2 + X + 1. Now, if we were to take this one half and multiply this back in, we will end up with the original X term, and adding the two constants together, we'll just cancel out the constants. So, this is going to be one way that we can manipulate this integral. Furthermore, what we can go ahead and do is we can pull out the 12 constant, and that is going to give us -16. In front of the integral. And if we were to split up this integral so that we can have the 2 X + 1 constant, or 2 X + 1 term in the numerator, we will end up with -16 multiplied by 2 X + 1 divided by X2 + X + 1, and we will have minus -16, which is going to give us 16, multiplied by 1 divided. By X2 + X + 1. Now, this 2nd integral that we've come up with here. Matches the form of the last integral that we have from the partial fraction decomposition. So we're going to combine these two integrals later. So let's just go ahead and focus on this first integral here. Now, by applying the U substitution, we can rewrite this integral to be -16. Multiplied by the integral of you. Sorry, 1 divided by UDU, and this is going to simplify to be -16 multiplied by the natural logarithm of U, which will backs substitute to be X2 + X + 1. So, this is going to be another value of the 2nd integral we are working with. Now let's go ahead and evaluate the 3rd integral. Now, for the 3rd integral, we have negative. 2 divided by 3 multiplied by X2 + X + 1. And if we were to add the 2nd integral that we evaluated earlier or came up with, we will end up getting plus the integral. And pushing the constant 1 divided by 6 multiplied by X2 + X plus 1. Now we can combine these integrals together if we were to add the least or multiply the first integral by 2 to both the numerator and denominator. That will leave us with negative integral of 4 divided by 6 multiplied by X2 + X + 1, plus the integral of 1 divided by 6 multiplied by X2 + X plus 1. We can combine these integrals together since they both have the same denominator, and that is going to combine to be negative 3, divided by 6 multiplied by X2 + X + 1. 3 and 6 were reduced to be 12, and if we were to pull that value to the front of the integral, we will be left with negative 1/2. Multiplied by the integral of 1 divided by X2 plus X plus 1. Now, we can simplify this integral if we were to complete the square and use the tangent inverse function. So if we were to complete the square on this integral, we will be left with negative. One half, multiplied by the integral of 1 divided by X. Plus 1/2 squared. Plus the square root of 3 divided by 2 squared. And if we were to use our trigonometric or trigonometric integral, which is the tangent inverse integral, this is going to evaluate to be negative, 1 divided by the square root of 3, multiplied by tangent inverse. Of 2 X plus 1 divided by the square root of 3. And this is going to be the final value for the 3rd integral. So, if we were to combine all three values together, the final solution to this problem is going to be 1/3, multiplied by the natural logarithm of X minus 1. Minus 16 multiplied by the natural logarithm of X2 + x + 1 minus 1 divided by the square root of 3, multiplied by the tangent inverse of 2 X plus 1 divided by the square root of 3, and because we were working with the with the indetermined integral, we are going to be adding a generic constant C, and this is going to be the final solution. To the original indefinite integral. And with that being said, the overall solution is going to be C. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

92–98. Evaluate the following integrals.
94. ∫ (dt / (t³ + 1))

Key Concepts

Integration

Partial Fraction Decomposition

Improper Integrals

Watch next