Video: Finding the General Antiderivative of a Function Using the Power Rule of Integration with Fractional Exponents

Determine ∫ (βˆ’4 ^(5)√(π‘₯⁹) + 8) ^(5)√(π‘₯Β²) dπ‘₯.

02:10

Video Transcript

Determine the indefinite integral having negative four multiplied by the fifth root of π‘₯ to the power of nine plus eight all multiplied by the fifth root of π‘₯ squared with respect to π‘₯.

Let’s start by writing this roots as powered. We know that the 𝑛th root of π‘₯ is equal to π‘₯ to the power of one over 𝑛. Once we’ve written our roots as powers, we can then combine them with the existing powers, using the fact that π‘₯ to the power of 𝑛 to the power of π‘š is equal to π‘₯ to the power of 𝑛 multiplied by π‘š. Therefore, π‘₯ to the power of nine to the power of one-fifth becomes π‘₯ to the power of nine-fifths. And π‘₯ squared to the power of one-fifths becomes π‘₯ to the power of two-fifths. Now, we can expand the brackets, using the fact that π‘₯ to the power of 𝑛 times π‘₯ to the power of π‘š is equal to π‘₯ to the power of 𝑛 plus π‘š. So our integral becomes the integral of negative four π‘₯ to the power of eleven-fifths plus eight multiplied by π‘₯ to the power of two-fifths with respect to π‘₯.

Here, we can use the power rule for integration which tells us that the indefinite integral of π‘₯ to the power of 𝑛 with respect to π‘₯ is equal to π‘₯ to the power of 𝑛 plus one over 𝑛 plus one plus 𝐢. We can apply this rule to our integral term by term. For the first term, we have negative four π‘₯ to the power of eleven-fifths. Therefore, 𝑛 is equal to eleven-fifths. When we integrate this term, we get negative four multiplied by π‘₯ to the power of 𝑛 plus one. And 𝑛 plus one is simply sixteen-fifths. So it’s π‘₯ to the power of sixteen-fifths. And then we need to divide by π‘š plus one. So that’s dividing by sixteen-fifths.

For the second time, we have eight multiplied by π‘₯ to the power of two-fifths. Therefore, 𝑛 is two-fifths. So we add eight multiplied by π‘₯ to the power of 𝑛 plus one which is π‘₯ to the power of seven-fifths. And we then divide by seven-fifths. And we mustn’t forget to add our constant of integration 𝐢. Now, all that remains to do is to simplify. And so we obtain a solution that the indefinite integral of negative four multiplied by the fifth root of π‘₯ to the power of nine plus eight all multiplied by the fifth root of π‘₯ squared with respect to π‘₯ is equal to negative five multiplied by π‘₯ to the power of sixteen-fifths over four plus 40 multiplied by π‘₯ to the power of seven-fifths over seven plus 𝐢.

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