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

Determine ∫(𝑥⁹ − 49𝑥⁷)/(𝑥⁸ − 7𝑥⁷) d𝑥.

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### Video Transcript

Determine the integral of 𝑥 to the ninth power minus 49𝑥 to the seventh power all divided by 𝑥 to the eighth power minus seven 𝑥 to the seventh power with respect to 𝑥.

In this question, we’re asked to evaluate the integral of a rational function. That’s the quotient of two polynomials. And it’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 𝑥 to the seventh power. And both terms in the denominator also share a factor of 𝑥 to the seventh power. So we’ll start by simplifying our integrand. First, we take the shared factor of 𝑥 to the seventh power out in our numerator, giving us a new numerator of 𝑥 to the seventh power multiplied by 𝑥 squared minus 49. And we’ll also take out the factor of 𝑥 to the seventh power in our denominator, giving us 𝑥 to the seventh power multiplied by 𝑥 minus seven.

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

And now we can cancel the shared factor of 𝑥 minus seven in the numerator and denominator. This then leaves us with the integral of 𝑥 plus seven with respect to 𝑥. 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 𝑎 and 𝑛, where 𝑛 is not equal to negative one, the integral of 𝑎𝑥 to the 𝑛th power with respect to 𝑥 is equal to 𝑎 multiplied by 𝑥 to the power of 𝑛 plus one divided by 𝑛 plus one plus the constant of integration 𝐶.

We can apply this to integrate each of the terms in our integrand separately. Let’s start with the first term. And it might be easier to rewrite this first term as 𝑥 to the first power. So the exponent of 𝑥 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 𝑥 squared over two. And we could add a constant of integration here. However, it’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 there’s a few different ways of doing this. For example, we could rewrite seven as seven 𝑥 to the zeroth power. We could then integrate this by using the power rule for integration. We add one to our exponent of 𝑥 and divide by this new exponent, giving us seven 𝑥 to the first power divided by one, which we can simplify to give us seven 𝑥. However, it’s much easier to remember, to integrate a constant, we just multiply it by 𝑥. This is because differentiating a constant multiple of 𝑥 with respect to 𝑥 just gives us the coefficient of 𝑥. The derivative of seven 𝑥 with respect to 𝑥 is equal to seven.

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