# Question Video: Finding the Integration of a Rational Function Involving Using Factorisation Mathematics • 12th Grade

Determine β«(π₯βΉ β 49π₯β·)/(π₯βΈ β 7π₯β·) dπ₯.

03:32

### Video Transcript

Determine the integral of π₯ to the ninth power minus 49π₯ to the seventh power all divided by π₯ to the eighth power minus seven π₯ to the seventh power with respect to π₯.

In this question, weβre asked to evaluate the integral of a rational function. Thatβs the quotient of two polynomials. And itβs very difficult to find the integral of a quotient of two functions. So we should start by trying to simplify our integrand.

To do this, we can notice both terms in our numerator share a factor of π₯ to the seventh power. And both terms in the denominator also share a factor of π₯ to the seventh power. So weβll start by simplifying our integrand. First, we take the shared factor of π₯ to the seventh power out in our numerator, giving us a new numerator of π₯ to the seventh power multiplied by π₯ squared minus 49. And weβll also take out the factor of π₯ to the seventh power in our denominator, giving us π₯ to the seventh power multiplied by π₯ minus seven.

We then want to integrate this with respect to π₯. And we can cancel the shared factor of π₯ to the seventh power in our numerator and denominator. However, we can see that weβre still integrating a rational function. So we should see if we can simplify this expression any further. In our numerator, we have π₯ squared minus 49. And we can notice this is a difference between squares. We know that π₯ squared is a square and 49 is seven squared. So π₯ squared minus 49 is a difference between squares. And we know that π squared minus π squared is equal to π minus π multiplied by π plus π. Therefore, we can rewrite this as the integral of π₯ minus seven multiplied by π₯ plus seven all divided by π₯ minus seven with respect to π₯.

And now we can cancel the shared factor of π₯ minus seven in the numerator and denominator. This then leaves us with the integral of π₯ plus seven with respect to π₯. This is just the integral of a polynomial. And we can do this by recalling the power rule for integration. We recall the power rule for integration tells us for any real constants π and π, where π is not equal to negative one, the integral of ππ₯ to the πth power with respect to π₯ is equal to π multiplied by π₯ to the power of π plus one divided by π plus one plus the constant of integration πΆ.

We can apply this to integrate each of the terms in our integrand separately. Letβs start with the first term. And it might be easier to rewrite this first term as π₯ to the first power. So the exponent of π₯ is one. We add one to this exponent to give a new exponent of two and then divide by this new exponent to give us π₯ squared over two. And we could add a constant of integration here. However, itβs easier to just add one constant of integration at the end of our expression.

We now need to integrate the second term in our integrand. And thereβs a few different ways of doing this. For example, we could rewrite seven as seven π₯ to the zeroth power. We could then integrate this by using the power rule for integration. We add one to our exponent of π₯ and divide by this new exponent, giving us seven π₯ to the first power divided by one, which we can simplify to give us seven π₯. However, itβs much easier to remember, to integrate a constant, we just multiply it by π₯. This is because differentiating a constant multiple of π₯ with respect to π₯ just gives us the coefficient of π₯. The derivative of seven π₯ with respect to π₯ is equal to seven.

So, using either method, weβve shown the integral of seven with respect to π₯ is seven π₯. And remember, we need to add a constant of integration πΆ. This then gives us our final answer. We were able to show the integral of π₯ to the ninth power minus 49π₯ to the seventh power all divided by π₯ to the eighth power minus seven π₯ to the seventh power with respect to π₯ is equal to π₯ squared over two plus seven π₯ plus a constant of integration πΆ.