Question Video: Finding the Integration of a Rational Function by Using Integration by Substitution | Nagwa Question Video: Finding the Integration of a Rational Function by Using Integration by Substitution | Nagwa

# Question Video: Finding the Integration of a Rational Function by Using Integration by Substitution Mathematics • Third Year of Secondary School

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Determine β«((π₯Β² + 7)/(π₯Β³ + 21π₯ β 5))ππ₯.

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### Video Transcript

Determine the indefinite integral of π₯ squared plus seven over π₯ cubed plus 21π₯ minus five with respect to π₯.

In order to solve this problem, we will be using integration by substitution. The way that we can spot this is if we take our denominator and call it π of π₯. So we have π of π₯ is equal to π₯ cubed plus 21π₯ minus five.

Now, we differentiate this to find π prime of π₯. Differentiating the π₯ cubed, we get three π₯ squared. Differentiating 21π₯, we get 21 and minus five differentiates to give zero. So π prime of π₯ is equal to three π₯ squared plus 21, which we can also write as three times by π₯ squared plus seven since we just factor out the three. And at this point, we notice that we have some constant which is three multiplied by the numerator of our fraction in the integral.

Now, we can rearrange this to write that π₯ squared plus seven is equal to one-third times π prime of π₯. So we can write our integral as one-third timesed by π prime of π₯ over π of π₯ with respect to π₯. And since that one-third is just a constant, we can factor out of the integral. And so that leaves us with this.

Now, our integral is of the form: the integral of πΉ of π of π₯ times π prime of π₯ with respect to π₯, where in our case the πΉ is πΉ of π of π₯ is equal to one over π of π₯. So now, our integral is of this form, we can use what is called a π’ substitution. And in this substitution, weβre going to let π’ equal π of π₯.

Now, weβre going to need to find ππ’ in terms of ππ₯. We can use the fact that ππ’ is equal to ππ’ by ππ₯ times ππ₯. Now, since π’ is equal to π of π₯, ππ’ ππ₯ is therefore equal to π prime of π₯. And so we get ππ’ is equal to π prime of π₯ times ππ₯.

We can rewrite our integral one-third lots of the integral times π prime of π₯ over π of π₯ ππ₯ as one-third lots of the integral of one over π of π₯ times π prime of π₯ with respect to π₯. And now, we notice that we have π prime of π₯ times π of π₯ which is the same as ππ’ of π₯.

So we are ready to make the substitution, remembering that π’ equals π of π₯ and ππ’ is equal to π prime times π of π₯. And so we get that this is equal to one-third times the integral of one over π’ ππ’. Now, all we need to do is calculate one-third times the integral of one over π’ ππ’. And we get that is equal to one-third times the natural logarithm of the absolute value of π’. Now, since this was an indefinite integral, we mustnβt forget to add on the constant of integration. So we get plus π on the end.

Next, we can do the inverse of our substitution. So we have that π’ is equal to π of π₯. So we can substitute this back in, giving us one-third times the natural logarithm of the absolute value of π of π₯ plus π. And for our final step, we can substitute back in that π of π₯ is equal to π₯ cubed plus 21π₯ minus five.

We get the final answer that the indefinite integral of π₯ squared plus seven over π₯ cubed plus 21π₯ minus five with respect to π₯ is equal to one-third times the natural logarithm of the absolute value of π₯ cubed plus 21π₯ minus five plus π.

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