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Determine the integral of negative four root 𝑥 minus five plus seven over 𝑥 squared d𝑥.
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So, the first thing we’re gonna want to do here is rewrite our expression if we want to integrate it.
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I’m gonna rewrite it using some exponent rules.
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And the first one of these is that if we’ve got the square root of 𝑥, then this is equal to 𝑥 raised to the power of a half.
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And we’re gonna be able to use this on our first term.
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And also, we’ve got another exponent rule that we’re gonna use.
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And that rule is if we have one over 𝑥 to the power of 𝑎, this is equal to 𝑥 to the power of negative 𝑎.
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And we’re gonna use that on the third term in our expression.
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Okay, so now, let’s use these and rewrite our expression.
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So, when we do, we can see that the expression we’re now trying to integrate is negative four 𝑥 to the power of a half minus five plus seven 𝑥 to the power of negative two.
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And in order to integrate the expression that we’ve got, we just need to remind ourselves how we integrate individual terms.
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Well, if we want to integrate 𝑥 to the power of 𝑛 d𝑥, then this is equal to 𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus 𝑐.
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So, basically, what we do is we add one to the exponent and divide by the new exponent.
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And then, we add 𝑐, which is our constant of integration.
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So, this means that our first term is gonna be negative four 𝑥 to the power of three over two.
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And that’s because if you have a half and you add a one, you get one and a half or three over two.
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And then, this is divided by the new exponent, so divided by three over two.
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Then, we get minus five 𝑥.
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And we get that because if we think of five being five 𝑥 to the power of zero, if we raise the exponent by one, we get five 𝑥.
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And if we divide by the new exponent, we divide by one; it doesn’t change it.
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And then, we get plus seven 𝑥 to the power of negative one over negative one.
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And then, finally, add 𝑐 because we can’t forget our constant of integration.
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Okay, so now, let’s tidy this up.
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Well, our first term is gonna be negative eight over three 𝑥 to the power of three over two.
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And if we look at how we got that, we had negative four divided by three over two.
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Well, if we divide by a fraction, it’s the same as multiplying by the reciprocal of that fraction.
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So, that’s negative four multiplied by two over three.
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Well, negative four multiplied by two is negative eight.
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So, we get negative eight over three.
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And then, our second term remains unchanged.
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So, we have negative five 𝑥.
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And then, finally, we have minus seven 𝑥 to the power of negative one.
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And we got that because it just changed the sign of the final term because we’re dividing by negative one.
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And then, we have our plus 𝑐 on the end.
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We could leave the answer like this.
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This will be totally acceptable.
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But we’re gonna rewrite it to be in the same format that we had the original expression in.
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Well, for the first term, what we’re gonna do is we’re going to use our exponent rule we looked at first.
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And what we’re gonna get is negative eight root 𝑥 cubed over three.
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So, we can use the first exponent rule to give us root 𝑥.
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But because we had 𝑥 to the power of three over two, then we use the numerator of that fraction as the exponent of the 𝑥 term within our root.
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And then, our second term remains unchanged.
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So, we’ve got minus five 𝑥.
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And then, finally, for the last term, we use the second exponent rule that we looked at.
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So, we’re gonna get minus seven over 𝑥.
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And then, we have our plus 𝑐 on the end.
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So, therefore, we can say that the integral of negative four root 𝑥 minus five plus seven over 𝑥 squared d𝑥 is negative eight root 𝑥 cubed over three minus five 𝑥 minus seven over 𝑥 plus 𝑐.