# Question Video: Finding the General Antiderivative of a Function Using the Power Rule of Integration with Fractional Exponents Mathematics • 12th Grade

Determine β« (β4 ^(5)β(π₯βΉ) + 8) ^(5)β(π₯Β²) dπ₯.

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### 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₯.

Letβs start by writing this roots as powered. We know that the πth root of π₯ is equal to π₯ to the power of one over π. Once weβve written our roots as powers, we can then combine them with the existing powers, using the fact that π₯ to the power of π to the power of π is equal to π₯ to the power of π multiplied by π. 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 π times π₯ to the power of π is equal to π₯ to the power of π plus π. 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 π with respect to π₯ is equal to π₯ to the power of π plus one over π plus one plus πΆ. 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, π is equal to eleven-fifths. When we integrate this term, we get negative four multiplied by π₯ to the power of π plus one. And π plus one is simply sixteen-fifths. So itβs π₯ to the power of sixteen-fifths. And then we need to divide by π plus one. So thatβs dividing by sixteen-fifths.

For the second time, we have eight multiplied by π₯ to the power of two-fifths. Therefore, π is two-fifths. So we add eight multiplied by π₯ to the power of π plus one which is π₯ to the power of seven-fifths. And we then divide by seven-fifths. And we mustnβt forget to add our constant of integration πΆ. 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 πΆ.