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Determine the integral of six 𝑥 plus eight over three 𝑥 squared plus eight 𝑥 plus three with respect to 𝑥.
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We will solve this problem using substitution.
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However, this will also lead us to a general rule that we can use for similar integration problems.
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Our first step is to let the denominator three 𝑥 squared plus eight 𝑥 plus three equal 𝑢.
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Differentiating this would give us d𝑢 by d𝑥.
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The differential of three 𝑥 squared is six 𝑥.
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And the differential of eight 𝑥 is eight.
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Differentiating the constant three gives us zero.
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Therefore, d𝑢 d𝑥 is equal to six 𝑥 plus eight.
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Rearranging this gives us that d𝑢 is equal to six 𝑥 plus eight d𝑥.
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We can now replace the denominator of our initial expression with 𝑢.
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And we can replace the numerator with d𝑢.
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We are therefore left with the integral of one over 𝑢 d𝑢.
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This is the same as one of our standard integrals that we should know.
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The integral of one over 𝑥 with respect to 𝑥 is equal to ln 𝑥 plus 𝑐.
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This means that the integral of one over 𝑢 with respect to 𝑢 will be ln 𝑢 plus 𝑐.
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Our final step is to substitute our value for 𝑢 back into the answer.
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The integral of six 𝑥 plus eight over three 𝑥 squared plus eight 𝑥 plus three is equal to ln of three 𝑥 squared plus eight 𝑥 plus three plus 𝑐.
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You might have noticed at the start that the top of our fraction was the differential of the bottom.
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Three 𝑥 squared plus eight 𝑥 plus three differentiated gives us six 𝑥 plus eight.
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If we are integrating 𝑓 dash 𝑥 over 𝑓 of 𝑥, our answer will always be the ln of 𝑓 of 𝑥.
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Another example that follows this method would be the integral of eight 𝑥 minus three over four 𝑥 squared minus three 𝑥 plus seven.
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Four 𝑥 squared minus three 𝑥 plus seven differentiates to eight 𝑥 minus three.
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This means that our answer would be 𝑙𝑛 of four 𝑥 squared minus three 𝑥 plus seven plus 𝑐.
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To integrate a quotient when the numerator is the derivative of the denominator, we can use this rule.