Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate ∫ 1/(π‘₯Β² βˆ’ 1)Β² dπ‘₯.

09:12

Video Transcript

Use partial fractions to evaluate the integral of the function one over π‘₯ squared minus one all squared with respect to π‘₯.

The first thing we need to do when we use partial fractions is to make sure the denominator of our fraction is factored fully. We can see that the π‘₯ squared minus one term in our denominator is the difference between two squares. So we can factorize this as π‘₯ minus one multiplied by π‘₯ plus one. So we can start by rewriting the fraction in our question as one over π‘₯ minus one multiplied by π‘₯ plus one all squared. We can simplify this term by evaluating the square around the parentheses. We can now use partial fractions to split this fraction into the following four fractions.

From our repeated root of π‘₯ minus one squared, we get two fractions of increasing powers of π‘₯ minus one. And from the repeated root of π‘₯ plus one squared, we get two fractions of increasing powers of π‘₯ plus one. At this point, we multiply both sides of our equation by the denominator π‘₯ minus one squared multiplied by π‘₯ plus one squared. We can cancel the shared factors on the left-hand side to give us one. We can then distribute the parentheses in the right-hand side of our equation. And finally, we can simplify this by canceling any shared factors in each of our fractions, giving us the following expression. Which we actually know is a conditional identity since it must be true for all values of π‘₯.

We might be tempted at this point to start equating coefficients of π‘₯ on both the left-hand and right-hand side of our equation. However, we know that this is true for all values of π‘₯. So we can simplify this problem by using substitution. We can notice that if we substitute in the value π‘₯ is equal to one, three of our four expressions will be zero. Substituting π‘₯ is equal to one gives us the following equation, where we notice that these three terms will be equal to zero.

We can simplify this to see that one is equal to 𝐡 multiplied by one plus one squared. Which we can rearrange to see that 𝐡 is equal to a quarter. We can now do the same by noticing that if we substitute the value of π‘₯ is equal to negative one, three of our terms again will be equal to zero. So substituting in the value of π‘₯ is equal to negative one will give us that one is equal to 𝐷 multiplied by negative one minus one squared. Which we can rearrange to find that 𝐷 must be equal to a quarter.

At this point, we’re going to compare the coefficients of the π‘₯-cubed terms on both sides of our equation. On the left-hand side of our equation, we have the constant one. So the coefficient of the π‘₯-cubed term must be zero. If we multiply out the parentheses on the right-hand side of our equation. We can see that the only terms which will have an π‘₯-cubed value are the ones with 𝐴 and 𝐢 as their coefficients. Giving us that when we equate the coefficients of the π‘₯-cubed terms, we must have that zero is equal to 𝐴 plus 𝐢. We can then get another equation involving 𝐴 and 𝐢 by comparing the constants on the left- and right-hand side of our equation. On the left-hand side of our equation, we just have the constant one. And on the right-hand side of our equation, we have that all four expressions will give us a constant. Which we can calculate by using the following expression.

Well, we must be careful to multiply by negative one wherever there is a negative one in the parentheses and to square any values which is squared within our parentheses. We can then simplify this to see that one is equal to negative 𝐴 plus 𝐡 plus 𝐢 plus 𝐷. We showed earlier that 𝐡 is equal to a quarter and 𝐷 is equal to a quarter. So we can substitute these values into this expression. We can then move all of the constants onto the left-hand side of this equation, giving us that one-half is equal to negative 𝐴 plus 𝐢. We can then add these two simultaneous equations together. Giving us that zero plus a half is equal to 𝐴 plus 𝐢 plus negative 𝐴 plus 𝐢. This then simplifies to give us that one-half is equal to two 𝐢. So we could see that 𝐢 must be equal to a quarter.

We can then use our equation that we found by comparing the coefficients of π‘₯ cubed, zero is equal to 𝐴 plus 𝐢, and the fact that 𝐢 is equal to a quarter. So we substitute 𝐢 is equal to a quarter into this equation. Giving us that 𝐴 must be equal to negative a quarter. We can now substitute the values of 𝐴, 𝐡, 𝐢, and 𝐷 into our original expression to find a new expression for our integrand. So let’s clear some space and rewrite what we have shown.

We have shown that we can rewrite the integrand in our question by using partial fractions in the following form. We can then integrate both sides of our expression with respect to π‘₯. On the right-hand side, we can split our integral into four separate integrals. And we can take out the constant coefficients in each of these. So now, we have four separate integrals to evaluate. At this point, we want to use one of our rules of integration which says that the integral of 𝑓 prime of π‘₯ divided by 𝑓 of π‘₯ with respect to π‘₯ is equal to the natural logarithm of the modulus of 𝑓 of π‘₯ plus our constant of integration.

We can see that in the integral of one over π‘₯ minus one with respect to π‘₯, if we let 𝑓 of π‘₯ equal π‘₯ minus one, then 𝑓 prime of π‘₯ is equal to one. So we arrive to the form of 𝑓 prime of π‘₯ over 𝑓 of π‘₯. So using our integral rule, we can evaluate this integral as the natural logarithm of the modulus of π‘₯ minus one plus a constant of integration 𝐢 one. Where we must be careful to multiply this by negative one-quarter because our integral was multiplied by negative a quarter. Similarly, in our integral of one over π‘₯ plus one with respect to π‘₯, if we let 𝑓 of π‘₯ equal π‘₯ plus one, we see that 𝑓 prime of π‘₯ is equal to one. Again, we can use our integral rule to evaluate this integral as the natural logarithm of the modulus of π‘₯ plus one plus our constant of integration 𝐢 two. Again, we multiply this by a quarter because our integral, which we were evaluating, was also multiplied by a quarter.

We now only have two integrals left to solve and we can solve both of these by using integration by substitution. We can start by letting 𝑒 equal π‘₯ minus one, giving us that d𝑒 dπ‘₯ must be equal to one. So we can say that d𝑒 is equal to dπ‘₯. We can then use all of this information to rewrite this integral as the integral of one over 𝑒 squared with respect to 𝑒. We know that one over 𝑒 squared is the same as 𝑒 to the power of negative two. And so, we can evaluate this integral by using our power rule for integrals, giving us negative 𝑒 to the power of negative one plus our constant of integration 𝐢 three. Finally, we substitute back in 𝑒 is equal to π‘₯ minus one.

We can now do the same with our final integral by using the substitution 𝑒 is equal to π‘₯ plus one. We can see that the derivative of 𝑒 with respect to π‘₯ will be equal to one. So we will have that d𝑒 is equal to dπ‘₯. Again, we can substitute this information into our integral giving us a quarter multiplied by the integral of one over 𝑒 squared with respect to 𝑒. We know that one over 𝑒 squared is the same as 𝑒 to the power of negative two. Next, we can evaluate this integral by using the power rule for integrals, giving us negative 𝑒 to the power of negative one plus our constant of integration 𝐢 four. Finally, we used the substitution 𝑒 is equal to π‘₯ plus one to give us the following expression.

Now, we just bring down our terms giving us the following expression which is equal to the evaluation of the integral in our question. We can then expand the parentheses and rearrange our equation to give us the following expression. We recall that the difference between two logarithms is equal to the logarithm of the quotient. So we can use this to rewrite the difference of the two natural logarithms in our expression as one single quotient of a natural logarithm. We can combine the two fractions in our expression by using cross multiplication, giving us the following expression. We can then cancel the shared factor of four. Then, we multiply out the parentheses in our numerator, giving us a new numerator of negative π‘₯ minus one minus π‘₯ plus one. We can then simplify the numerator to just be equal to negative two π‘₯.

We can then cancel the shared factor of two. And finally, we can multiply our two factors in the denominator to get a denominator of two multiplied by π‘₯ squared minus one. Finally, since 𝐢 one, 𝐢 two, 𝐢 three, and 𝐢 four are all constants, the combination of all of these together will just be a constant, which we will call π‘˜. Giving us a final answer that the integral in our question can be evaluated by using partial fractions to be one-quarter multiplied by the natural logarithm of the modulus of π‘₯ plus one over π‘₯ minus one minus π‘₯ over two multiplied by π‘₯ squared minus one plus a constant π‘˜.

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