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

Determine β«((β2π₯ β 5)Β²/π₯)ππ₯.

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Video Transcript

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

First, letβs expand the numerator of the fraction in the integral. So thatβ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, letβ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 weβ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 mustnβt forget that since this is an indefinite integral, we need to add on the constant of integration. So we get plus π 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 π.