Original subtitles

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In this video we're doing another partial fractions problem and we've been asked to find the integral

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of one divided by x cubed plus x.

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Now at first glance this might not look like a partial fractions problem because it doesn't look like

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we have any factors in our denominator.

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But what we have to realize is that this denominator can actually be factored.

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So we can actually call this the integral of one divided by here in the denominator if we factor out

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an x we get X times x squared plus 1 D X..

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So now we have factors in our denominator we have one linear factor this x to the first power factor

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here and we have one quadratic factor the x squared plus 1.

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Remember when you you have X to the First it's a linear factor when you have x squared.

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It's a quadratic factor so to set up our partial fraction decomposition will take this original fraction

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here with the denominator factored and we'll put that on the left hand side of this new equation.

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So X times quantity x squared plus 1 on the right hand side.

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Remember whenever you have a linear factor you just have a single constant in the numerator and then

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just add these fractions together.

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We have our quadratic factor X squared plus 1.

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And whenever you have a quadratic factor your numerator needs to look like and B X plus C.

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We start with B because A is already taken.

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So now this is going to be our partial fractions decomposition.

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This value over here on the right hand side.

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This is the value that we're going to use to replace the original function.

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So we're going to eventually be taking the integral of this instead of the original fraction.

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All we need to do first is find values for A B and C..

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So how are we going to find values for A B and C Well as always we want to multiply both sides by the

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denominator from the left hand side.

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So that denominator is X times quantity x squared plus 1.

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So we want to multiply every term by these two factors.

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When we do that on the left hand side over here we're going to get this X to cancel with this X and

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we're going to get X squared plus 1 to cancel this x squared plus 1.

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So we're just going to be left with one on the left hand side on the right hand side.

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When we multiply x 10 x squared plus 1 by a divided by X we're going to get this X to cancel with this

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factor of x leaving us with just this x squared plus 1.

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So we get a times quantity x squared plus 1.

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And then here when we multiply these two factors by this be X possi fraction we're going to get the

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x squared plus 1 in the denominator to cancel with this x squared plus 1.

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Leaving us with just that X there.

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So we're going to get plus B X plus C multiplied by x.

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Now what we want to do is multiply out the right hand side.

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So we're going get one is equal to a x squared plus a plus b x squared plus C x.

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Now keep in mind that over here on the left hand side we just have one which is a constant.

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So we could really change this left hand side and we could write 0.

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X squared plus 0 x plus 1.

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That original Cosson we haven't changed anything by adding 0 X because 0 times X is 0.

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We also haven't changed anything by adding 0 x squared because zero times x square it is still 0.

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So we've just added 0 to the left hand side.

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But this will help us visualize our method of undetermined coefficients that we're going to go to in

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a second.

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So we have that left hand side over here on the right hand side we want to group together like terms

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we're going to pull all of our x squared terms together.

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So we're going to say a x squared plus B X squared to take care of these 2 x squared terms.

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Then we want all of our x terms together.

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In this case let's just see X right there and then we want all of our constants together.

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And in this case that's just a.

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So that deals with that term.

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Then on the right hand side we want to factor out the x value wherever we can.

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So here we're going to pull x squared out of X squared plus B x square which is just going to leave

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us with quantity a plus be multiplied by x squared.

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Then we're going to have this.

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See times X and then plus a.

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Now the reason we pulled out the X variables is because at this point we can go ahead and equate coefficients

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from the left in the right hand side.

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In other words we know that the coefficient on x squared on the right hand side is a plus b right here.

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The coefficient on x squared.

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On the left hand side is zero which means we can say A plus B is equal to zero.

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We also know here that the coefficient on X on the right is c and of the coefficient on X on the left

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is zero.

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So we can say C equals zero and we have a constant of a on the right and a constant of one on the left.

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So we can say A is equal to 1.

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So when we set up those three equations Here's what we get.

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WE HAVE A plus B equals zero.

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We have C equals zero and we have a equals one our goal in doing this is to solve for the three values

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here a b and c so we can take this value in the purple box and evaluate the integral of this function

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instead of the integral of the original function.

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So we already have a in C we know A's equal to one in C's equal to zero to find a B.

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We're just going to take this equation plug in equals one and instead of A plus B equals zero we get

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1 plus B is equal to zero.

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Subtracting one from both sides we get B is equal to negative 1.

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So now we plug those three values into this right hand side here instead of a will put in 1 instead

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of B we'll put in negative 1.

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And instead of C we'll put in zero and now we want to rewrite our integral.

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So our integral looks like this instead of the original integral we'll say 1 over X so 1 over X we have

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a plus negative 1 times X we'll go ahead and say plus negative x and then we have plus 0 so don't need

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to write that so we have negative X divided by x squared plus 1 D x.

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Now we can split this into two integrals we can say the integral of one over x d x will pull this minus

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sign out and we'll say minus the integral of x over x squared plus 1 D x.

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Now for this second integral here this first one is easy but for the second one we need to use substitution.

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So I will say you is equal to the denominator x square plus 1.

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We'll say d u the derivative of you is equal to 2 x d x and then we'll solve for x by dividing both

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sides by 2 x and we'll get x is equal to do you over to x.

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So now will evaluate this first integral we know that the integral of 1 divided by X is the natural

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log of x are going to get natural laga of the absolute value of x according to this formula here.

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Then for our second integral I will say minus the integral leave this X in the numerator the denominator

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x squared plus 1 we said equal to use we have you there.

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And then we know dx is d u over to X so we have do you over to x.

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Now we know that we can get X to cancel from the numerator and denominator.

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So what we end up with is natural log the absolute value of X minus we'll pull this two from the denominator

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out in front so we'll say minus 1 1/2 times the integral of 1 over u d u.

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Well we know that the integral of one over you is natural log of the absolute value of use we get natural

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log absolute value of X minus 1 1/2 times the natural log of the absolute value of you.

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And then we add c to account for the constant of integration.

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We know that you is x squared plus once we just have the back substitute and then we can say that our

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final answer is the natural log of the absolute value of X minus 1 1/2 times the natural log of the

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absolute value of x squared plus 1 plus C and this value here is our final answer.

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It's the value of the original integral which we found using partial fractions and distinct quadratic

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factors.