Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate ∫ 1/(𝑑³ + 𝑑² βˆ’ 2𝑑) d𝑑.

03:10

Video Transcript

Use partial fractions to evaluate the indefinite integral of one over 𝑑 cubed plus 𝑑 squared minus two 𝑑 with respect to 𝑑.

When rewriting the integrand using partial fractions, we’ll be looking to reverse the process we would take when adding algebraic fractions. Before we can do this though, we need to decide what the denominator of these fractions might be. So let’s see if we can factor the denominator.

It’s clearly a cubic, but there’s a common factor of 𝑑. So we’ll factor the denominator by 𝑑 to get 𝑑 times 𝑑 squared plus 𝑑 minus two. We can then factor this quadratic expression the usual way. 𝑑 squared plus 𝑑 minus two is equal to 𝑑 minus one times 𝑑 plus two. We can now split this into partial fractions. These are 𝐴 over 𝑑 plus some other constant 𝐡 over 𝑑 minus one plus 𝐢 over 𝑑 plus two.

Next, we’re going to add these three fractions. We’ll need to multiply the numerator and denominator of each of them by the denominator of the other two. So the numerator becomes 𝐴 times 𝑑 minus one times 𝑑 plus two plus 𝐡 times 𝑑 times 𝑑 plus two plus 𝐢 times 𝑑 times 𝑑 minus one. And now we notice that the denominator of the fractions on both sides of this equation are equal. And for the fractions themselves to be equal, this means in turn that their numerators must be equal. So we can say that one is equal to 𝐴 times 𝑑 minus one times 𝑑 plus two plus 𝐡𝑑 times 𝑑 plus two plus 𝐢𝑑 times 𝑑 minus one. We’re going to use the method of substitution to help us find the values for 𝐴, 𝐡, and 𝐢.

Notice that if we let 𝑑 be equal to zero, we can instantly eliminate 𝐡 and 𝐢. So we substitute 𝑑 equals zero into our equation. And we find that one is equal to 𝐴 times zero minus one times zero plus two, which simplifies to one equals negative two 𝐴. Dividing both sides of this equation by negative two, and we obtain 𝐴 to be equal to negative one-half.

Next, we spot that if we let 𝑑 be equal to one, we’re going to eliminate 𝐴 and 𝐢. Substituting 𝑑 equals one gives us one is equal to three 𝐡. And when we divide by three, we see that 𝐡 is equal to one-third. Finally, we’ll let 𝑑 be equal to negative two. This will eliminate 𝐴 and 𝐡 this time. This gives us one is equal to six 𝐢. And dividing through by six, we find 𝐢 to be equal to one-sixth.

Let’s clear some space for the next step. We can say that one over 𝑑 times 𝑑 minus one times 𝑑 plus two is equal to negative one over two 𝑑 plus one over three times 𝑑 minus one plus one over six times 𝑑 plus two. We’re now ready to integrate with respect to 𝑑. The integral of one over 𝑑 is the natural log of the absolute value of 𝑑. So our first term integrates to negative a half times the natural log of 𝑑. The interval of one over three times 𝑑 minus one is a third times the natural log of the absolute value of 𝑑 minus one. And the integral of one over six times 𝑑 plus two is equal to a sixth times the natural log of the absolute value of 𝑑 plus two. We mustn’t forget since we’re dealing with an indefinite integral to add that constant of integration 𝐢.

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