Original subtitles

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In this video we're talking about partial fractions and how to use partial fractions to evaluate an

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integral and in this particular problem we're going to be dealing with distinct linear factors and we

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call it distinct linear factors because when you look at the denominator you have three linear factors

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x x plus two and X minus 1 So those are the three linear factors we call them linear because they all

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involve X to the first powers.

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So this is X to the first is X to the first power plus two is X to the first power minus one of x to

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the first power.

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It's a linear factor.

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And we call them distinct because they aren't equal to each other.

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So a factor of x is not the same as a factor of x plus 2 which is not the same as a factor of x minus

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1.

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This is the easiest kind of partial fractions problem distinct linear factors because linear factors

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are easier to handle than quadratic factors and distinct factors are easier to handle than repeated

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factors.

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So when you have distinct linear factors this is how you do the partial fractions decomposition you're

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going to keep the original function the original fraction here exactly as is on the left hand side so

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we're going to say for x squared minus 3 X minus 4 divided by the original denominator X times x plus

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two times X minus 1.

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We're going to set that equal to.

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And we're going to give each factor its own fraction.

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So this X is going to be in one fraction then we're going to put the X plus two in the next fraction

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and the X minus one in the next fraction.

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And then when you have linear factors the numerators are just a single constant.

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So a.

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And then because we already used a we used B and then because we've already used a and b we use.

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See this right hand side is what we're going to use to replace the original fraction here.

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So we're actually going to end up integrating this instead of the original fraction.

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All we have to do before we get to that step is find values for A B and C..

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So how do we go about finding those values for the constants A B and C.

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Well we're going to multiply both sides of this equation by the denominator from the left hand sides

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are going multiply everything by x times x plus two times X minus 1.

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When we do that we're going to get the denominator from the left hand side to go away completely that

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that'll cancel with this value.

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So we're left with four x squared minus three X minus four.

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And then on the right hand side this X here will cancel with this X leaving us with just x plus 2 times

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X minus 1.

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So we'll get a times quantity x plus two times quantity x minus 1 for this fraction here involving b.

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We'll get the X plus two to cancel with this X plus two.

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Leaving us with just X and X minus once we'll say plus B times x times quantity x minus 1.

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And then for our last fraction involving C We'll get the X Minus One to cancel with this X Minus One

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leaving us with just c times x times x plus 2.

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Now we want to do is go ahead and expand the right hand side we'll leave the left hand side alone for

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right now on the left hand side here we're going to get a time's quantity x plus two times quantity

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x minus one is going to give us X squared minus X plus 2 x or just Plus X and then minus 2.

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Here we're going to get B X squared minus B X..

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When we distribute the B X across the X minus one.

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And then here we're going to get c x squared plus to see X..

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Lastly we'll distribute the a across the quantity x squared plus X minus 2.

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And so will get a x squared plus a X minus 2 A and the rest will be the same.

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Now we want to grouped together like terms so again we'll leave the left hand side as is for right now

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on the right hand side we want to put all of our x squared terms together.

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So that's going to look like a x squared plus and B X squared plus C x squared.

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That's going to cover this term this term and this term then we want to put all of our first degree

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x terms together.

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So we'll put these in parentheses here then we'll say plus a X minus B X plus 2.

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C x.

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So that's going to cover this this and this.

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And then lastly we'll put all of our constants together.

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So we're going to end up with a minus to a and I'll cover the last her there.

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Next we'll factor out the x variable from each of these sets here of parentheses.

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So again we leave the left hand side and on the right hand side we'll pull out an x squared.

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So that'll just leave us with quantity A plus B plus C when we factor in x squared out of X squared

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plus square pussy x squared.

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And then here will factor out an x.

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So we'll be left with a minus B plus 2 C and then we pull that X out and then we have the Today.

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Now the reason that we do it like this is because we want to equate coefficients from the left and the

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right hand side of the equal sign.

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So you can do is draw boxes around these coefficients to make it even more clear for yourself.

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So what we see is that the coefficient on the x squared term on the right hand side is this a plus b

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plus C that we created the coefficient on x squared.

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On the left hand side is this value here.

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So we're going to be able to do is say four is equal to a plus b plus C.

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We can also say that a minus B plus 2 C is going to be equal to negative 3 because those are the coefficients

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on the first degree x variable.

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And then lastly the constants were going to be able to say negative 2A is going to be equal to negative

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4.

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Since those are the constants on each side.

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So we want to set up those equations for equals a plus b plus C negative 3 is equal to a minus B plus

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2 C and then a negative 4 is equal to negative to a.

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Well the easiest one to solve obviously is this negative 4 equals negative 2.

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So we'll divide both sides by negative two and we'll get positive two is equal to a.

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So a is equal to two.

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We'll put that up here and then give ourselves more room.

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So if a is equal to to what we can say then when we plug a was 2 into both these equations is instead

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of four equals a plus be Plessey we'll get four equals two plus B plus see if we subtract two from both

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sides then we'll end up with two is equal to B plus C.

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Same thing here with this equation will get negative three equals two minus B plus to see if we subtract

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two from both sides we're going to get negative three minus two is a negative five equals negative B

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plus 2.

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See now we can do is add these equations together because if we add them together we're going to get

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B plus a negative b which is zero.

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So to minus five or two plus a negative five gives us a negative three equals C plus 2 C gives us 3

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C.

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And if you divide both sides by three you get C is equal to negative 1.

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Now if we take that value for C and the plug it back into this first equation here instead of two equals

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B plus C we get two equals B minus 1.

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And then if we add 1 to both sides we add B is equal to 3.

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So now we have the values equal to be equal 3 and C equal negative 1.

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We're going to be taking this original partial fractions decomposition this right hand side that we

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found here and we're going to be plugging that in for the original fraction that we started with.

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So we're going to be replacing this original fraction with the partial fractions the composition except

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that instead of a B and C we're going to have two three and negative one.

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So instead of taking the integral of this original fraction we're going to say the integral of two over

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X plus three over X plus two and then we have a negative one so minus 1 over X minus 1 D X and since

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we know that the integral of one over x is equal to natural log of the absolute value of x plus C we

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can go ahead and say then that our integral is to we take this from the numerator times.

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Now we have two times one over X so two times the natural log of the absolute value of x.

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Here we're going to have plus three times the natural log of the absolute value of x plus 2.

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And here we're going to have minus 1 so minus natural log of the absolute value of X minus 1.

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And then we add c to account for our cost of integration and that's how you use partial fractions to

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evaluate an integral that has distinct linear factors.