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

Determine ∫((−2𝑥 − 5)²/𝑥)𝑑𝑥.

03:32

### Video Transcript

Determine the indefinite integral of minus two 𝑥 minus five all squared over 𝑥 with respect to 𝑥.

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

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

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

Now, let’s look at how to integrate each of these terms individually. The first two terms here four 𝑥 plus 20 are polynomial. So we can simply integrate them by increasing the 𝑥 power by one and dividing by the new power. So for the four 𝑥 term, we get four 𝑥 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 𝑥 to the nought since 𝑥 to the nought is just one. So when we increase the 𝑥 power by one, we get 20 times 𝑥 to the one. And now, we divide by the new power, which is one.

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

And we’ve just written that the integral of one over 𝑥 with respect to 𝑥 is equal to the natural logarithm of the absolute value of 𝑥. So therefore, this is equal to 25 times the natural logarithm of the absolute value of 𝑥. 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 𝑥 minus five all squared over 𝑥 with respect to 𝑥 is equal to two 𝑥 squared plus 20𝑥 plus 25 times the natural logarithm of the absolute value of 𝑥 plus 𝑐.