Video: Finding the Integration of a Rational Function by Distributing the Division

Determine ∫((βˆ’2π‘₯ βˆ’ 5)Β²/π‘₯)𝑑π‘₯.

03:32

Video Transcript

Determine the indefinite integral of minus two π‘₯ minus five all squared over π‘₯ with respect to π‘₯.

First, let’s expand the numerator of the fraction in the integral. So that’s minus two π‘₯ minus five all squared. And this is equal to minus two π‘₯ minus five times minus two π‘₯ minus five. Multiplying these together, we get four π‘₯ squared plus 10π‘₯ plus 10π‘₯ plus 25, which simplifies to four π‘₯ squared plus 20π‘₯ plus 25.

Now, we can write this back into our integral. And this gives us the indefinite integral of four π‘₯ squared plus 20π‘₯ plus 25 all over π‘₯ with respect to π‘₯. Now, we can split this fraction up. Now, we have the integral of four π‘₯ squared over π‘₯ plus 20π‘₯ over π‘₯ plus 25 over π‘₯ with respect to π‘₯.

Now we notice that in the first two fractions, we have π‘₯ is on both the numerator and denominator. So we can cancel out these π‘₯s. This leaves us with the indefinite integral of four π‘₯ plus 20 plus 25 over π‘₯ with respect to π‘₯.

Now, let’s look at how to integrate each of these terms individually. The first two terms here four π‘₯ plus 20 are polynomial. So we can simply integrate them by increasing the π‘₯ power by one and dividing by the new power. So for the four π‘₯ term, we get four π‘₯ squared. And the new power is two. So we must divide by two.

Now, the second term is 20. And we can write this as 20 times π‘₯ to the nought since π‘₯ to the nought is just one. So when we increase the π‘₯ power by one, we get 20 times π‘₯ to the one. And now, we divide by the new power, which is one.

Now the third term, 25 over π‘₯, is a little more difficult to calculate. We must remember the rule that the integral of one over π‘₯ with respect to π‘₯ is equal to the natural logarithm of the absolute value of π‘₯. However, in our case, we have the integral of 25 over π‘₯ with respect to π‘₯. Since 25 is just a constant, we can factor out of the integral and write 25 times the integral of one over π‘₯ with respect to π‘₯.

And we’ve just written that the integral of one over π‘₯ with respect to π‘₯ is equal to the natural logarithm of the absolute value of π‘₯. So therefore, this is equal to 25 times the natural logarithm of the absolute value of π‘₯. So now, we can add this term on. And finally, we mustn’t forget that since this is an indefinite integral, we need to add on the constant of integration. So we get plus 𝑐 on the end.

Now, we have found our solution. We just need to simplify it. We get a final solution that the indefinite integral of minus two π‘₯ minus five all squared over π‘₯ with respect to π‘₯ is equal to two π‘₯ squared plus 20π‘₯ plus 25 times the natural logarithm of the absolute value of π‘₯ plus 𝑐.

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