Video: Integration of Rational Functions by Partial Fractions

Use partial fractions to evaluate ∫_(1/2) ^(1) (π‘₯ + 4)/(π‘₯(π‘₯ + 1)) dπ‘₯.

04:22

Video Transcript

Use partial fractions to evaluate the integral from a half to one of π‘₯ plus four divided by π‘₯ multiplied by π‘₯ plus one with respect to π‘₯.

The question wants us to use partial fractions to evaluate the integral given to us in the question. We notice that our faction has two unique roots in the denominator. Since the denominator of our fraction has two unique factors, we can use partial fractions to represent it as some constant 𝐴 divided by π‘₯ plus some constant 𝐡 divided by π‘₯ plus one. We’ll simplify this by multiplying both sides of the equation by the denominator π‘₯ multiplied by π‘₯ plus one. This gives us that π‘₯ plus four is equal to 𝐴 multiplied by π‘₯ plus one plus 𝐡π‘₯.

This equation is true for all values of π‘₯. We’ll represent this by using an equivalent sign. We can eliminate 𝐴 from this equation by setting π‘₯ equal to negative one. We can also eliminate 𝐡 from our equation by setting π‘₯ equal to zero. Substituting π‘₯ is equal to negative one gives us negative one plus four is equal to 𝐴 multiplied by negative one plus one plus 𝐡 multiplied by negative one. This simplifies to give us that three is equal to negative 𝐡. So we multiply both sides of our equation by negative one. We see that 𝐡 is equal to negative three.

Now we substitute π‘₯ is equal to zero to get zero plus four is equal to 𝐴 multiplied by zero plus one plus 𝐡 multiplied by zero. Which we can simplify to give us that four is equal to 𝐴. So by using partial fractions, we’ve shown that we can represent our integrand of π‘₯ plus four divided by π‘₯ multiplied by π‘₯ plus one. As four over π‘₯ minus three divided by π‘₯ plus one.

We’re now ready to evaluate our definite integral. We’ll start by changing our integrand by using our partial fraction representation. Next, we’ll split our integral into two separate integrals. We recall for a constant 𝐴. The integral of 𝐴 divided by π‘₯ with respect to π‘₯ is equal to 𝐴 multiplied by the natural algorithm of the absolute value of π‘₯ plus the constant of integration 𝑐. We can use this to evaluate our first integral. The integral from a half to one of four divided by π‘₯ with respect to π‘₯ is equal to four. Multiplied by the natural logarithm of the absolute value of π‘₯ evaluated between π‘₯ is equal to a half and π‘₯ is equal to one. Well, we don’t include our constant of integration because we’re calculating a definite integral.

To calculate our second integral, we’re going to use the substitution 𝑒 is equal to π‘₯ plus one. Differentiating both sides of this equation with respect to π‘₯ gives us that the derivative of 𝑒 with respect to π‘₯ is equal to one. Giving us the equivalent statement d𝑒 is equal to dπ‘₯.

Finally, since we’re using substitution on a definite integral, we need to calculate the new limits of our integral. For our upper limit, we have when π‘₯ is equal to one 𝑒 is equal to π‘₯ plus one, which is one plus one, which is equal to two. And for our lower limit, when π‘₯ is equal to a half, we have that 𝑒 is equal to a half plus one is equal to three over two.

So by using the substitution 𝑒 is equal to π‘₯ plus one. Our second integral becomes the integral from three over two to two of negative three over 𝑒 with respect to 𝑒. And we see this is now in the form of our integral rule. So we can use this to evaluate the integral.

Evaluating our second integral gives us negative three multiplied by the natural logarithm of the absolute value of 𝑒. Between the limits of 𝑒 is equal to three over two and 𝑒 is equal to two. We’re now ready to evaluate the limits of our integral. This gives us four multiplied by the natural logarithm of the absolute value of one minus four. Multiplied by the natural logarithm of the absolute value of a half minus three. Multiplied by the natural logarithm of the absolute value of two plus three. Multiplied by the natural logarithm of the absolute value of three over two.

We know that the absolute value of one is one. The absolute value of a half is a half. The absolute value of two is two. And the absolute value of three over two is three over two. So we can remove all of our absolute values.

Next, we notice that the natural logarithm of one is just equal to zero. After rearranging this, we have shown that, by using partial fractions, we can evaluate the integral from a half to one of π‘₯ plus four divided by π‘₯ multiplied by π‘₯ plus one with respect to π‘₯. To be equal to negative three multiplied by the natural logarithm of two minus four. Multiplied by the natural logarithm of a half plus three multiplied by the natural logarithm of three over two.

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