Original subtitles

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

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partial fractions is just an integration technique similar to use substitution or integration by parts.

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It's just a different way to evaluate an integral based on the kind of function that we have.

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Usually when you have a fraction like this when we are five divided by quantity to X plus one times

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quantity x minus two.

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When you have something like that partial fractions might be a good way to evaluate the integral What

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part of fractions allows us to do is to simplify a fraction like this one that has factors in its denominator

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such that we come up with a function that's easier to integrate than the function that we've been given.

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So all we want to do is recognize first of all what kind of factors we have in our denominator in the

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denominator of this fraction.

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We have two factors One factor to X plus one.

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The other factor is X minus 2.

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So we need to first distinguish between linear factors and quadratic factors.

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These are both linear factors because this is actually 2 x to the first power.

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And this is X to the first power in other words we have first degree X variables with X to the first.

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So they're linear factors if you have X squared or a larger degree exponent on the x variable you're

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looking at a quadratic factor linear factors are going to be easier for us to handle than quadratic

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vectors.

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The other thing we look at is whether or not we have distinct factors or repeated factors.

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In other words do we have multiple of the same factor or all of our factors different in this case 2

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x plus one is different than X minus 2.

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So these two factors are distinct from one another.

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They're not the same.

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So we have distinct linear factors.

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And when that's the case here's what your partial fractions decomposition is going to look like you're

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going to keep the fraction exactly as it is we're going to put that on the left hand side of a new equation

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that we're going to create.

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So we're going to say five divided by quantity to X plus one times quantity x minus two so exactly the

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same fraction that we started with.

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Then on the right hand side we're going to put each of the linear factors in the denominator of its

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own fraction.

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So 2 x plus 1 is going to go in the denominator of one fraction and X minus two is going to go in the

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denominator of another fraction and then because these are linear factors we're going to put a single

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constant in the numerator of each fraction.

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So we'll call this constant a.

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And then because we've already used a we'll call this constant B.

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Now this right hand side over here this is what we're going to use instead of the original fraction

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that we were given.

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So we're going to take this value here and we're going to replace this original fraction with this right

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hand side.

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The only thing we have to do first is solve for values of A and B the way that we're going to find values

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for A and B is by multiplying both sides of this equation by the denominator from the left hand side.

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So we're going to multiply everything by quantity to X plus one times quantity x minus two this denominator

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here from the left hand side.

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When we do that we're gonna get this entire denominator to cancel so we're just going to be left with

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five on the left hand side on the right hand side.

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We'll get this to X plus 1 in the denominator to cancel with this 2 x plus 1.

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Leaving us with just the factor of x minus 2.

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So we're going to get a times quantity x minus 2.

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And then when we multiply this by the second fraction here with B we're going to get X minus 2 in the

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denominator to cancel with this X minus 2 in the numerator and we're going to get B times 2 x plus 1.

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Now there's a couple ways that we can solve for the values of A and B at this point.

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We can multiply everything out on the right hand side collect like terms and then use the method of

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undetermined coefficients.

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But when the facts are this simple we can use another method we can say if x is equal to 2 then inside

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these parentheses here we're going to get to minus 2 or 0 0 times a is 0.

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In other words setting X equal to two is going to get rid of this constant.

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A It's going to get rid of the variable A and we're just going to be left with B.

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So we'll be able to solve for a value of be.

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So you want to pick the value of x that's going to make the value inside the parentheses 0.

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So here if we say x is equal to 2 so 4 x equals to what we can get is five is equal to.

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Here we get to minus 2 or zero zero times a is zero.

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So that term disappears.

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Over here we're going to get B times 2 times 2 is 4 4 plus 1 is 5.

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So we're going to get the times 5 If we divide both sides by 5 We're going to get B is equal to 1.

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Now we can do the same thing.

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But to solve for a.

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We just need to make this 2 x plus 1 equal to zero.

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Because if we make this equal to zero then we multiply it by B we'll get B to disappear.

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So how do we make 2 x plus one equals zero.

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Well this one's a little more complicated but we can just say to X plus one equals zero and solve for

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x so 2 x is equal to negative 1 x is equal to negative 1 1/2.

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So if we say for X equals negative 1 half what we get then is 5 is go to a multiplied by negative one

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half minus two is a negative 5 1/2.

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So we get negative five halves here we're going to get plus B when we take negative one having plug

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it in here for X. We're going to get negative 1 1/2 times 2 is negative one negative one plus one is

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0 0 times b is zero.

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So this whole term disappears.

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Now if we multiply here both sides by 2 what we get is 10 is equal to negative five.

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And then if we divide both sides by negative 5 we get a is equal to negative 2.

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So now we have values for a and b we can plug them into our partial fraction decomposition.

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So we'll say a is negative 2 and we'll say B is going to be 1.

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And now we can take this value and replace the original fraction.

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So instead of taking the integral of this original fraction Instead we'll take the integral of negative

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2 over 2 x plus one plus one over X minus 2 D x.

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And now this is an integral we can easily evaluate.

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Remember we know that the integral of 1 divided by X is the natural log of the value of x plus C.

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In other words and we want to take the integral of 1 divided by something whatever this is here the

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integral of that is going to be natural log of the absolute value of that denominator plus C.

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The only thing we have to remember is that we have to apply chain rule so we'll have to divide this

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by the derivative of x.

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So here's what that looks like.

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We can separate these two fractions into their own integrals and we can pull out this constant here.

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So I end up with is negative two times the integral of one divided by two x plus 1 D x plus the integral

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of 1 over X minus 2 D x.

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Now when we take the integral we can take this negative to put that out in front so negative 2 then

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we'll say the natural log of the absolute value of the denominators of the natural log of the absolute

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value of the denominator to X plus 1.

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But then we have to divide it by the derivative of the denominator.

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So our denominator is 2 x plus 1 the derivative of 2 x plus 1 is just 2.

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So we have to divide by 2.

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Then for this integral here the second integral will say plus the natural log of the absolute value

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of x minus 2.

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But now we have to divide by the derivative of the denominator.

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So the denominator is X minus 2.

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The derivative of X minus 2 is just 1.

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So we have to divide this by 1 but of course dividing by 1 doesn't change anything so we don't actually

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have to write that in.

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And then we just put plus C to account for that constant of integration.

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So then here we can cancel the two in the numerator with the two in the denominator.

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Leave me with just negative natural log of the value of 2 x plus 1.

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If we reorder these terms we get natural log absolute value X minus 2 minus natural Lague absolute value

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of 2 x plus 1 plus C.

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And remember when you have natural log of a minus natural log of B.

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That's going to be equal to the natural log of a divided by B.

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So what we can say here is we can say that this is equal to the natural log of the absolute value of

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this first value.

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Here this first argument.

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X minus 2 divided by the second argument to X plus 1 and then we can add our constant integration.

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So this then is our final answer.

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This is the integral of the original function which we found using a partial fraction decomposition

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with distinct linear factors.