### Video Transcript

Determine the integral of negative two π₯ over π₯ plus eight squared with respect to π₯.

We can solve this integration problem by substitution. Our first step is to let π’ equal π₯ plus eight. This means that the denominator will become π’ squared. If π’ is equal to π₯ plus eight, then dπ’ dπ₯, the differential, is equal to one. Whilst the dπ’ and dπ₯ technically canβt be split, for the purposes of integration by substitution, we can now write that dπ’ is equal to dπ₯. This enables us to eliminate dπ₯ from the initial integration. The final thing we need to eliminate is the π₯ on the numerator. As π’ is equal to π₯ plus eight, then π₯ will be equal to π’ minus eight.

We can now substitute in these values so we have an integration in terms of π’. The numerator becomes negative two multiplied by π’ minus eight. The denominator is π’ squared. As dπ’ was equal to dπ₯, we can now integrate this with respect to π’. Distributing the parentheses gives us negative two π’ plus 16. We can now split the numerator so that we have two terms that we can integrate individually. The first term becomes negative two π’ over π’ squared. We can then cancel our π’, leaving us with negative two over π’. The second term becomes 16 over π’ squared.

Weβre now going to integrate negative two over π’ and 16 over π’ squared with respect to π’ individually. At this point, it is worth recalling a couple of our key laws or rules of integration. If weβre integrating the constant π over π₯ with respect to π₯, this is equal to π ln of π₯ plus π. The integral of negative two over π’ is therefore negative two ln π’. For our second term, it is worth rewriting 16 over π’ squared as 16π’ to the power of negative two.

We can then use our general rule for integrating that the integral of ππ₯ to the power of π with respect to π₯ is equal to ππ₯ to the power of π plus one over π plus one plus π. The integral of 16π’ to the power of negative two is therefore 16π’ to the power of negative one over negative one. Dividing a positive number by a negative number gives us a negative. Therefore, this simplifies to negative 16π’ to the power of negative one. This in turn can be rewritten as negative 16 over π’. We now need to return to our initial substitution, π’ equals π₯ plus eight, and substitute this back in. This gives us negative two ln of π₯ plus eight minus 16 over π₯ plus eight. Our last step is to add our constant of integration π. We must do this every time weβre dealing with an indefinite integral.

The integral of negative two π₯ over π₯ plus eight all squared with respect to π₯ is negative two ln of π₯ plus eight minus 16 over π₯ plus eight plus π. Note that we use the absolute value or modulus when dealing with any logs.