### Video Transcript

Determine the integral of six π₯ plus eight over three π₯ squared plus eight π₯ plus three with respect to π₯.

We will solve this problem using substitution. However, this will also lead us to a general rule that we can use for similar integration problems. Our first step is to let the denominator three π₯ squared plus eight π₯ plus three equal π’. Differentiating this would give us dπ’ by dπ₯. The differential of three π₯ squared is six π₯. And the differential of eight π₯ is eight.

Differentiating the constant three gives us zero. Therefore, dπ’ dπ₯ is equal to six π₯ plus eight. Rearranging this gives us that dπ’ is equal to six π₯ plus eight dπ₯. We can now replace the denominator of our initial expression with π’. And we can replace the numerator with dπ’. We are therefore left with the integral of one over π’ dπ’. This is the same as one of our standard integrals that we should know. The integral of one over π₯ with respect to π₯ is equal to ln π₯ plus π. This means that the integral of one over π’ with respect to π’ will be ln π’ plus π.

Our final step is to substitute our value for π’ back into the answer. The integral of six π₯ plus eight over three π₯ squared plus eight π₯ plus three is equal to ln of three π₯ squared plus eight π₯ plus three plus π. You might have noticed at the start that the top of our fraction was the differential of the bottom. Three π₯ squared plus eight π₯ plus three differentiated gives us six π₯ plus eight. If we are integrating π dash π₯ over π of π₯, our answer will always be the ln of π of π₯.

Another example that follows this method would be the integral of eight π₯ minus three over four π₯ squared minus three π₯ plus seven. Four π₯ squared minus three π₯ plus seven differentiates to eight π₯ minus three. This means that our answer would be ππ of four π₯ squared minus three π₯ plus seven plus π. To integrate a quotient when the numerator is the derivative of the denominator, we can use this rule.