Question Video: Finding the Integration of a Rational Function by Using Integration by Substitution | Nagwa Question Video: Finding the Integration of a Rational Function by Using Integration by Substitution | Nagwa

# Question Video: Finding the Integration of a Rational Function by Using Integration by Substitution Mathematics • Third Year of Secondary School

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Determine β« β2π₯/(π₯ + 8)Β² dπ₯.

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### 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 canβt 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.

Weβre 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 weβre 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 ππ₯ to the power of π with respect to π₯ is equal to ππ₯ 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 weβre 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.

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