If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:8:52

try to evaluate the following integral so I'm assuming you've had a go at it so let's let's work through this together and if at any point you get inspired always feel free to pause the video and continue on with it on your own so the first thing that might have jumped out at you we have a rational expression the degree in the numerator is the same as the degree in the denominator so maybe a little bit of algebraic long division is called for so let's do that let's take x squared minus 1 let's take x squared minus 1 and divide it into x squared within a different color divide it into x squared plus X minus 5 so x squared plus X minus 5 and so let's look at the highest degree term so how many times as x squared go into x squared well it goes one time let me write this in a new color goes one time one times x squared minus one is just going to be x squared minus one and I subtract this green expression from this move expression or I could just add the negative on it so let me just take the negative of it so it's going to be negative x squared plus one and we're going to get the X Squared's x squared minus x squared is zero so those cancel out and we're going to be left with X we're going to be left with X and the negative five plus one is negative four so we have X minus four left over so we can rewrite the expression that we're trying to find the antiderivative of we can rewrite it as one one plus X minus four one plus X minus four over over x squared minus one over x squared maybe we'll do that in that purple color since I already used it as the purple over x squared minus one so we did one thing we now we have a lower degree in the numerator than we have in the denominator and obviously this is fairly straightforward to take the antiderivative of but what do we do now it's not clear if we look at x squared minus one its derivative would be two X which is the same degree as this but it's not X minus four so doesn't look like you substitution is going to is going to help us with this so what can we do now and now we can take out another tool in our algebraic toolkit is we can sense you do or we will do partial fraction expansion which is essentially writing this as the sum of two rational expressions that have a lower degree in the denominator so what do I mean by that so this term right over here X minus four over x squared minus one we can rewrite that as X minus four over instead of x squared minus one we can factor this this is X plus one times X minus one so let's write that this is X plus one times X minus one and when we think about partial fraction expansion we say okay well can we write this as the sum of as the sum of something let's call that a over over X plus one over X plus one plus something else let's call that B plus B over X minus one can we do that and to attempt to do that let's if we had to just add these two things what would we get well we would find a common denominator which would be X plus one times X minus one and so you would have and if any of this looks from unfamiliar I encourage you to review the Witte videos on partial fraction expansion because that's exactly what we're doing right over here but this would be equal to if you were to add the two your common denominator would be the product so it'd be X plus one it would be X plus 1 times X minus 1 X minus 1 and then so the first term I would multiply the numerator and the denominator times X minus 1 so it'd be a times X minus 1 a times X minus 1 plus B the second term are multiplied the numerator denominator times X plus 1 times X plus 1 and so what do we get we this is going to be equal to this is going to be equal to ax maybe I'll do this all in one color this is going to be equal to a whoops a X minus a plus B X plus B plus B X plus B and then all of that over the stuff that we keep writing over and over again actually let me just copy and paste this so copy and paste I could use that over and over again so we have that over that let's see so now we can group the X terms and so we can rewrite this as so if we take ax plus BX that's going to be a plus B times X and then we have and then we have a negative a and a B so plus B minus a and I'll just put parentheses around that just so I kind of group these these constant terms and then all of that's going to be divided by good thing I copied and pasted that X plus 1 times X minus 1 and so now this is the this is the crux of partial fraction expansion we say okay we kind of went through this whole exercise on the thesis that we could do this that there is some a and B for which this is true and so if there is some a and B for which this is true then a plus B must be the coefficient on the X term right over here so a plus B must be equal to 1 must be equal to this coefficient and B minus a must be equal to the constant must be equal to negative 4 or if they are then we will found an A and a B so let's do that so I'll do it up here since I have a little bit of real estate so a plus B is going to be equal to 1 and B minus a or I could write that as negative a plus B is equal to negative 4 is equal to negative 4 C we could add the left-hand sides and add the right-hand sides and then we would the A's would disappear we would get 2 B is equal to negative 3 or B is equal to negative 3 halves and then a is we know that a is equal to a is equal to one minus B which would be equal to one plus three-halves since b is negative three-halves which is equal to five halves a is equal to 5 halves b is equal to negative three halves and just like that we can rewrite this whole integral in a way that is a little bit easier to take the antiderivative 1/1 plus a over X plus one a is 5 halves and so I could just write that is let me write it this way five halves times one over X plus one and I wrote it that way because it's very straightforward to take the antiderivative of this and then plus B over X minus one which is going to be negative three halves so I'll just write it as minus three halves minus three halves times one over X minus one that was this right over here DX DX notice all I did is I took this I took this expression right over here and I did a little bit of partial partial fraction expansion into these into these two I guess you could say expressions or terms right over there and now it's fairly straightforward to integrate this antiderivative of 1 is just going to be X antiderivative of five halves 1 over X plus one it's going to be plus five halves the natural log of the absolute value of x plus one we were able to do that because the derivative of x plus 1 is just 1 so the derivative is there so that we can take the antiderivative with respect to X plus 1 you could also do u substitution like we've done in previous examples U is equal to X plus one and over here this is going to be minus three-halves times the natural log of the absolute value of x minus 1 by the same exact logic where how we would take the antiderivative there and of course we cannot forget our constant and there we have it we've been able to integrate we were able to evaluate you ate this expression