Video: Finding the Integration of a Function Involving Using Factorisation

Determine ∫(π‘₯⁴ βˆ’ 14π‘₯Β³)/(π‘₯ βˆ’ 14) dπ‘₯.

02:18

Video Transcript

Determine the integral of π‘₯ to the fourth power minus 14π‘₯ cubed over π‘₯ minus 14 with respect to π‘₯.

Now, at first glance, this integral looks a little bit complicated because we’re being asked to integrate a quotient of two polynomial functions. But if we look more closely, we’ll see that, in fact, the denominator of this quotient is a factor of the numerator. π‘₯ to the fourth power minus 14π‘₯ cubed can be factors as π‘₯ cubed multiplied by π‘₯ minus 14. And so, what we see is that, actually, this factor of π‘₯ minus 14 in the numerator will entirely cancel the denominator out.

So, cancelling these two factors, and we’re just left with the integral of π‘₯ cubed over one, or π‘₯ cubed with respect to π‘₯, a far more straightforward integral to perform. To do so, we need to recall the power rule of integration. This tells us that the integral of π‘₯ to the power of 𝑛 with respect to π‘₯, for some real number 𝑛 not equal to negative one, is equal to π‘₯ to the power of 𝑛 plus one over 𝑛 plus one plus a constant of integration 𝑐 if we’re performing an indefinite integral. In practical terms, what this means is we increase the exponent by one and then divide by the new exponent.

Hopefully, you can see why this rule can’t be applied when 𝑛 is equal to negative one. If we did follow this rule, we’d have π‘₯ to the power of zero over zero. And division by zero is undefined. So, we need an alternative approach if the exponent we’re trying to integrate is negative one. Our exponent though is three, so we can go ahead and apply this rule. Increasing the exponent by one gives π‘₯ to the fourth power. And then, dividing by this new exponent gives one-quarter π‘₯ to the fourth power. We must remember to add a constant of integration 𝑐 as we’re performing an indefinite integral.

So, by first simplifying the function we were asked to integrate by factoring, and then applying the power rule of integration, we found that the indefinite integral of π‘₯ to the fourth power minus 14π‘₯ cubed over π‘₯ minus 14 with respect to π‘₯ is one-quarter π‘₯ to the fourth power plus a constant of integration 𝑐.

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