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Determine the indefinite integral of π₯ squared plus seven over π₯ cubed plus 21π₯ minus five with respect to π₯.
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In order to solve this problem, we will be using integration by substitution.
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The way that we can spot this is if we take our denominator and call it π of π₯.
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So we have π of π₯ is equal to π₯ cubed plus 21π₯ minus five.
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Now, we differentiate this to find π prime of π₯.
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Differentiating the π₯ cubed, we get three π₯ squared.
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Differentiating 21π₯, we get 21 and minus five differentiates to give zero.
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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.
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And at this point, we notice that we have some constant which is three multiplied by the numerator of our fraction in the integral.
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Now, we can rearrange this to write that π₯ squared plus seven is equal to one-third times π prime of π₯.
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So we can write our integral as one-third timesed by π prime of π₯ over π of π₯ with respect to π₯.
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And since that one-third is just a constant, we can factor out of the integral.
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And so that leaves us with this.
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Now, our integral is of the form: the integral of πΉ of π of π₯ times π prime of π₯ with respect to π₯, where in our case the πΉ is πΉ of π of π₯ is equal to one over π of π₯.
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So now, our integral is of this form, we can use what is called a π’ substitution.
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And in this substitution, weβre going to let π’ equal π of π₯.
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Now, weβre going to need to find ππ’ in terms of ππ₯.
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We can use the fact that ππ’ is equal to ππ’ by ππ₯ times ππ₯.
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Now, since π’ is equal to π of π₯, ππ’ ππ₯ is therefore equal to π prime of π₯.
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And so we get ππ’ is equal to π prime of π₯ times ππ₯.
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We can rewrite our integral one-third lots of the integral times π prime of π₯ over π of π₯ ππ₯ as one-third lots of the integral of one over π of π₯ times π prime of π₯ with respect to π₯.
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And now, we notice that we have π prime of π₯ times π of π₯ which is the same as ππ’ of π₯.
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So we are ready to make the substitution, remembering that π’ equals π of π₯ and ππ’ is equal to π prime times π of π₯.
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And so we get that this is equal to one-third times the integral of one over π’ ππ’.
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Now, all we need to do is calculate one-third times the integral of one over π’ ππ’.
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And we get that is equal to one-third times the natural logarithm of the absolute value of π’.
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Now, since this was an indefinite integral, we mustnβt forget to add on the constant of integration.
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So we get plus π on the end.
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Next, we can do the inverse of our substitution.
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So we have that π’ is equal to π of π₯.
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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.
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And for our final step, we can substitute back in that π of π₯ is equal to π₯ cubed plus 21π₯ minus five.
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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.