Question Video: Integration by Partial Fractions of the Rational Function

Use partial fractions to evaluate the integral ∫ (2π‘₯ + 3)/(10π‘₯Β³ + 21π‘₯Β² + 9π‘₯) dπ‘₯.

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Video Transcript

Use partial fractions to evaluate the integral of two π‘₯ plus three divided by 10π‘₯ cubed plus 21π‘₯ squared plus nine π‘₯ with respect to π‘₯.

The question gives us an integral, and it wants us to rewrite this integral by using partial fractions. To use partial fractions, we’re going to need to fully factor the denominator of our integrand. To do this, we start by noticing each term shares a factor of π‘₯. So we take this factor out. We get π‘₯ times 10π‘₯ squared plus 21π‘₯ plus nine. We now need to factor this quadratic. We see that our leading coefficient is 10. And 10 is equal to five times two. So we’ll try to factor our quadratic as five π‘₯ plus something times two π‘₯ plus something.

We can then see that nine is equal to three times three. So if we try this in our factors, we get five π‘₯ plus three times two π‘₯ plus three. And we can then see that five π‘₯ times two π‘₯ is 10π‘₯ squared, and three times three is equal to nine. Finally, we can see that three times two π‘₯ plus five π‘₯ times three is equal to 21. In other words, we can factor our denominator as π‘₯ times five π‘₯ plus two times two π‘₯ plus three. Now, we’ll use this factorization to rewrite our integral. We get the integral of two π‘₯ plus three divided by π‘₯ times five π‘₯ plus three times two π‘₯ plus three with respect to π‘₯.

And we can now see that in our integrand, our numerator and our denominator share a factor of two π‘₯ plus three. So we’ll just cancel these out. This then gives us the integral of one divided by π‘₯ times five π‘₯ plus three with respect to π‘₯. And now, we’ll write our integrand of one divided by π‘₯ times five π‘₯ plus three by using partial fractions. We can see the denominator of our integrand has two unique factors. So by partial fractions, we can rewrite this as 𝐴 over π‘₯ plus 𝐡 divided by five π‘₯ plus three for some constants 𝐴 and 𝐡. And to find these values of 𝐴 and 𝐡, we’re going to multiply both sides by π‘₯ times five π‘₯ plus three.

Doing this, we get that one is equivalent to 𝐴 times five π‘₯ plus three plus 𝐡 times π‘₯. And remember, this is true for all values of π‘₯. So if we substitute π‘₯ is equal to zero, we’ll eliminate our variable 𝐡. Substituting π‘₯ is equal to zero, we get one is equal to 𝐴 multiplied by five times zero plus three plus 𝐡 times zero. We have five times zero plus three is equal to three and 𝐡 times zero is equal to zero. So this simplifies to give us one is equal to three 𝐴. And then we can just divide through by three. We get one-third is equal to 𝐴.

To find the value of 𝐡, we’re going to substitute π‘₯ is equal to negative three divided by five. This will eliminate our variable 𝐴. Substituting in π‘₯ is equal to negative three over five, we get one is equal to 𝐴 times five multiplied by negative three over five plus three plus 𝐡 times negative three over five. We can then simplify this expression. We get one is equal to negative three 𝐡 over five. Then we’ll multiply both sides of this equation by negative five over three. We see that 𝐡 is equal to negative five over three.

So by using partial fractions, we’ve shown we can rewrite one over π‘₯ times five π‘₯ plus three as 𝐴 over π‘₯ plus 𝐡 over five π‘₯ plus three. And we’ve shown the value of 𝐴 will be equal to one-third. And we also showed that the value of 𝐡 will be negative five divided by three. We can then use this to rewrite our integral. We get the integral of one over three π‘₯ minus five divided by three times five π‘₯ plus three with respect to π‘₯. We’ll rewrite our integral of the difference between two functions as the difference of their integrals. This gives us the integral of one over three π‘₯ with respect to π‘₯ minus the integral of five over three times five π‘₯ plus three with respect to π‘₯.

Next, we’ll take our constant factors out of these integrals. We’ll take a constant factor of one-third out of our first integral and the factor of five over three out of our second integral. This gives us one-third times the integral of one over π‘₯ with respect to π‘₯ minus five over three times the integral of one over five π‘₯ plus three with respect to π‘₯. To evaluate our first integral, we recall the integral of the reciprocal function with respect to π‘₯ is equal to the natural logarithm of the absolute value of π‘₯ plus a constant of integration 𝐢.

To evaluate our second integral, we’ll do this by using a 𝑒 substitution. We’ll use 𝑒 equal to our denominator five π‘₯ plus three. Differentiating both sides with respect to π‘₯, we get d𝑒 by dπ‘₯ is equal to five. And then, this gives us the equivalent statement in terms of differentials d𝑒 is equal to five times dπ‘₯. Then, if we take a look at the coefficient of our second integral, we can take five back inside of our integral. This gives us a term of five dπ‘₯, which we know is equal to d𝑒.

So applying our integral rule and our 𝑒 substitution, we get one-third times the natural logarithm of the absolute value of π‘₯ plus a constant of integration we’ll call 𝐢 one minus one-third times the integral of one over 𝑒 with respect to 𝑒. And of course, we can evaluate the integral of one over 𝑒 with respect to 𝑒 by using our integral law. Evaluating our integral of one over 𝑒 with respect to 𝑒, distributing one-third over both sets of our parentheses. We get one-third times the natural logarithm of the absolute value of π‘₯ plus 𝐢 one over three minus one-third times the natural logarithm of the absolute value of 𝑒 minus our constant of integration 𝐢 two divided by three.

And we can simplify this expression. First, 𝐢 one divided by three minus 𝐢 two divided by three is a constant, so we’ll just call this 𝐾. This gives us the following expression. Next, remember, we’re calculating the integral of a function of π‘₯, so we want to give our answer in terms of π‘₯. We’ll do this by using our substitution 𝑒 is equal to five π‘₯ plus three. This gives us one-third times the natural logarithm of the absolute value of π‘₯ minus one-third times the natural algorithm of the absolute value of five π‘₯ plus three plus 𝐾.

Now, we’ll take out the shared factor of one-third from our first two terms. Doing this, we get one-third times the natural logarithm of the absolute value of π‘₯ minus the natural logarithm of the absolute value of five π‘₯ plus three plus our constant 𝐾. Now, we want to simplify this by using one of our logarithm laws. The logarithm of π‘š minus the logarithm of 𝑛 is equal to the logarithm of π‘š divided by 𝑛. This would give us the natural logarithm of the absolute value of π‘₯ divided by the absolute value of five π‘₯ plus three. But instead of dividing the absolute value of two expressions, we can just take the absolute value of the entire expression.

So we can replace this with the natural logarithm of the absolute value of π‘₯ divided by five π‘₯ plus three. And this gives us our final answer. Therefore, by using partial fractions, we’ve shown the integral of two π‘₯ plus three divided by 10π‘₯ cubed plus 21π‘₯ squared plus nine π‘₯ with respect to π‘₯. Is equal to one-third times the natural logarithm of the absolute value of π‘₯ divided by five π‘₯ plus three plus a constant of integration 𝐾.

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