Partial Fractions

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Think about this integral: x plus one, over x minus two times x minus three, d x. No standard form matches it, right? But here is a secret: that fraction is exactly minus three over x minus two, plus four over x minus three. And each piece is just a log! So where did that minus three and that four come from? Today we learn to break a fraction into simple pieces, the method of partial fractions. It comes up often in JEE problems, so let's begin.

0:40First check: is the fraction proper?

So, partial fractions work on a rational function: P of x over Q of x, where P and Q are polynomials. But first, one check. Okay? Take the fraction on screen, x cubed over x squared minus one. Can we split it into partial fractions straight away? No, we cannot split it yet; we must look at the degrees first. The short word deg means degree, and the sign after it is greater than or equal to. If the degree of the numerator is not less than that of the denominator, divide first. Students often skip this check, so watch out, okay? Divide x cubed by x squared minus one: quotient x, remainder x. So x cubed over x squared minus one equals x, plus x over x squared minus one. The leftover fraction is proper now, and only this part gets split. Clear?

1:36One fraction for each factor

Now, once it is proper, factor the denominator; each factor gets its own fraction. For a linear factor, x minus a, we write capital A over x minus a, where capital A is a constant to be found. Simple, right? But what if a factor repeats, say x minus a, whole squared? Is one fraction, capital A over x minus a whole squared, enough? No, it is not enough, and this is a very common trap. So, a repeated factor, x minus a to the power k, needs k fractions, one for every power from one up to k. The subscripts below each A just label them, and the three dots stand for the terms in between. Got it? Third, a quadratic with no real linear factors; we call it irreducible. Over it we need a linear top: B x plus C, over x squared plus p x plus q. Careful, this C is just a number to be found, not the constant of integration. With me?

2:42The cover-up rule

Now, to find these constants fast for distinct linear factors, the ones that do not repeat, use the cover-up rule. Capital A equals P of x, over Q of x over x minus a, at x equals a. The slash means over, and the tall bar with x equals a at its foot means: put x equals a. Okay? In plain words: cover the factor x minus a in the denominator, and put x equals a in everything that is left. That is all there is to it. Here is why it works: multiply both sides by x minus a, and put x equals a. Every other fraction becomes zero, and only capital A is left. Right?

3:27One more trick: fractions in eˣ and sin x

One more trick, for fractions built from e to the power x, or sine x, and so on. We cannot factor these like polynomials. With me? Instead, write the numerator as capital A times the denominator, plus capital B times the derivative of the denominator; the dot means times. Then match the two sides to get capital A and capital B, okay? Now think. After this split, what does each part integrate to? The first part is just capital A, giving capital A x. The second is the derivative of the bottom over the bottom, so it gives capital B times natural log of mod of the denominator. The l n is the natural log, and the straight bars are the modulus. Plus capital C, the constant. Clear?

4:18Example 1: x + 1 over (x - 2)(x - 3) by the cover-up rule

Now, remember our puzzle? Let's finish it. Example one. Find the integral of x plus one, over x minus two times x minus three, d x. The long S is the integral sign, and d x means with respect to x. Pause the video and try it first. Done? Okay, let's check. Degree one over degree two, so it is proper. Two distinct linear factors, so we write capital A over x minus two, plus capital B over x minus three. Okay? Now the cover-up rule. Cover x minus two, and put x equals two in what is left. What do you get for capital A? Correct, capital A is minus three! Two plus one is three on top, and two minus three is minus one below. Right? The same way for capital B: cover x minus three, and put x equals three. Three plus one is four, over three minus two, which is one. So capital B is four. There they are, the minus three and the four from our puzzle! Now each piece is a log. The integral is minus three times natural log of mod of x minus two, plus four times natural log of mod of x minus three, plus capital C, the constant of integration. And our puzzle is solved. See how quick that was?

5:49Example 2: 1 over (x + 1)(x2 + 1), Bx + C over the quadratic

Now a harder one, Example two. Find the integral of d x, over x plus one, times x squared plus one. One linear factor and one quadratic factor. Pause the video and try it first. Done? Okay, let's check. It is proper, and x squared plus one has no real linear factors, so it is irreducible. So what goes on top of it? Just a constant? No, careful here! Over an irreducible quadratic we need a linear top. So: capital A over x plus one, plus B x plus C, over x squared plus one. Okay? Now x plus one does not repeat, so cover-up still works. Cover x plus one and put x equals minus one: one over, minus one squared plus one, which is one over two. So capital A is one half. For capital B and capital C, multiply through by the full denominator. One equals capital A times x squared plus one, plus B x plus C, times x plus one. With me? Now compare coefficients. The left side has no x squared term. So what is capital B? Yes, exactly, capital B is minus one half! Capital A plus capital B is zero, so capital B is minus capital A; the double arrow on screen means so. And the constant terms give capital A plus capital C equals one, so capital C is one half. Got it? Split the second fraction: minus one half times x over x squared plus one, plus one half times one over x squared plus one. Look at the first part: the top, x, is half the derivative of x squared plus one. Right? So x over x squared plus one gives one half natural log of x squared plus one; times minus one half, that makes minus one quarter. No modulus here, since x squared plus one is always positive. And one over x squared plus one gives tan inverse x; the little minus one means inverse, not a power. Answer: one half natural log of mod of x plus one, minus one quarter natural log of x squared plus one, plus one half tan inverse x, plus capital C, the constant of integration, a new one, not the C we found above. Clear?

8:22Quick revision: divide first, fraction per factor, cover-up

So, quick revision. Check degrees first: if the top is not smaller, divide. Then one fraction per factor, one for every power of a repeated factor, and B x plus C over an irreducible quadratic. For distinct linear factors, use cover-up. Then integrate each piece, and add plus capital C, okay? Now go and try the practice problems on this page. All the best!