Use partial fractions to evaluate
the indefinite integral of 𝑥 plus four over 𝑥 plus six times 𝑥 minus one with
respect to 𝑥.
Remember, we need to rewrite this
integrand using partial fractions. So we begin by reversing the
process we would take when adding algebraic fractions. We write it as 𝐴 over 𝑥 plus six
plus 𝐵 over 𝑥 minus one. And there are two ways for us to
work out the constants 𝐴 and 𝐵. They’re substitution and equating
Let’s begin by looking at the
substitution method. Let’s imagine we’re adding these
fractions. We multiply the numerator and
denominator of the first fraction by 𝑥 minus one and the numerator and denominator
of the second fraction by 𝑥 plus six. So we see that 𝑥 plus four over 𝑥
plus six times 𝑥 minus one is equal to 𝐴 times 𝑥 minus one plus 𝐵 times 𝑥 plus
six all over 𝑥 plus six times 𝑥 minus one.
Notice that the denominator of the
fractions on both sides of our equation are equal. This means that, for the fractions
themselves to be equal, their numerators must be. And we can say that 𝑥 plus four is
equal to 𝐴 times 𝑥 minus one plus 𝐵 times 𝑥 plus six. Okay, so far, so good.
Now we want to find a way to
eliminate one of the constants from this equation. Well, we see that if we set 𝑥 to
be equal to one, this bit here becomes 𝐴 times one minus one, which is 𝐴 times
zero, which is zero. So let’s set 𝑥 be equal to
one. When we do, we see that one plus
four is equal to 𝐴 times zero plus 𝐵 times one plus six, which simplifying gives
us five equals seven 𝐵. And now we have an equation in
So we can solve this by dividing
both sides by seven. And we see that 𝐵 is equal to
five-sevenths. Brilliant, so let’s repeat this
process to help us establish the value of 𝐴. And you might wish to pause the
video for a moment and think about what substitution would eliminate 𝐵 from this
If we let 𝑥 be equal to negative
six, then the second term over here becomes 𝐵 times zero. So the 𝐵’s going to be
eliminated. And if we let 𝑥 be equal to
negative six, our entire equation becomes negative six plus four equals 𝐴 times
negative six minus one plus 𝐵 times zero, which simplifies to negative two equals
negative seven 𝐴. Now we have an equation in 𝐴. And dividing both sides by negative
seven, we obtain 𝐴 to be equal to two-sevenths. And we’ve successfully decomposed
into partial fractions.
We can say that 𝑥 plus four over
𝑥 plus six times 𝑥 minus one is equal to two over seven times 𝑥 plus six plus
five over seven times 𝑥 minus one. And we’re now of course able to
integrate our expression with respect to 𝑥. Remembering that the integral of
the sum of functions is the same as the sum of the integrals of those respective
functions. And of course, we can take out any
constant factors. And we see that our integral is
equal to two-sevenths of the integral of one over 𝑥 plus six d𝑥 plus five-sevenths
of the integral of one over 𝑥 minus one d𝑥.
Well, the integral of one over 𝑥
plus six is the natural log of the absolute value of 𝑥 plus six. And the integral of one over 𝑥
minus one is the natural log of the absolute value of 𝑥 minus one. And of course, since this is an
indefinite integral, we must add that constant 𝑐. And we’re done. The indefinite integral of 𝑥 plus
four over 𝑥 plus six times 𝑥 minus one with respect to 𝑥 is two-sevenths times
the natural log of the absolute value of 𝑥 plus six plus five-sevenths of the
natural log of the absolute value of 𝑥 minus one plus the constant 𝑐.
We are now going to consider how we
could’ve got here using the method of equating coefficients. The starting process is the
same. We need to get to this stage
here. We want to distribute the
parentheses on the right-hand side. And when we do, we see that 𝑥 plus
four is equal to 𝐴𝑥 minus 𝐴 plus 𝐵𝑥 plus six 𝐵.
Now this next step isn’t entirely
necessary. But it can help us figure out what
to do next. We collect together like terms. And we see that 𝑥 plus four is
equal to 𝐴 plus 𝐵 times 𝑥 plus negative 𝐴 plus six 𝐵 or six 𝐵 minus 𝐴. And now we have two families of
terms, if you will. We have 𝑥 to the power of ones,
and then we have these constants. And we can say that those are 𝑥 to
the power of zeros.
We want to equate coefficients for
these terms. Let’s begin by equating the
coefficients of 𝑥 to the power of zero, or just the constants. On the left-hand side, we have
four. And on the right-hand side, we have
six 𝐵 minus 𝐴. Next, we’ll equate the coefficients
of 𝑥 to the power of one. The coefficient of 𝑥 on the
left-hand side is one. And on the right-hand side, that’s
𝐴 plus 𝐵. So we now have a pair of
simultaneous equations which we can begin to solve by first adding.
Negative 𝐴 plus 𝐴 is zero. So we see that when we add our pair
of simultaneous equations, we end up with five equals seven 𝐵. And solving this equation for 𝐵,
we find that 𝐵 is equal to five-sevenths. We’re gonna then substitute this
value of 𝐵 into either of our original equations. I’m going to choose this one
here. So one is equal to 𝐴 plus
five-sevenths. Then subtracting five-sevenths from
both sides, we obtain 𝐴 to be equal to two-sevenths. And the rest of the process is
exactly the same.
We have our partial fractions, and
we can integrate each of them. And we obtain the indefinite
integral to be equal to two-sevenths of the natural log of 𝑥 plus six plus
five-sevenths of the natural log of 𝑥 minus one plus 𝑐.