# Question Video: Finding the Integration of a Function Using the Power Rule for Integration with Roots and Negative Exponents Mathematics • 12th Grade

Determine β«(8β(π₯) β (1/π₯β΄)) dπ₯.

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### Video Transcript

Determine the integral of eight root π₯ minus one divided by π₯ to the fourth power with respect to π₯.

The question wants us to integrate eight times the square root of π₯ minus one divided by π₯ to the fourth power. This integrand is in a form which we donβt yet know how to integrate. So weβre going to need to manipulate this expression into one which we can integrate. Since both of these expressions involve exponents, weβll try rewriting this by using our laws of exponents. The first thing we notice is the first term in our integrand contains the square root of π₯. And one of our laws of exponents tells us we can rewrite the square root of π₯ as π₯ to the power of one-half. And we know how to integrate π₯ to the power of one-half by using the power rule for integration.

Next, by using our laws of exponents, we can rewrite one divided by π₯ to the fourth power as π₯ to the power of negative four. And again, we can integrate this by using the power rule for integration. So, by using our laws of exponents, weβve rewritten our integral as the integral of eight π₯ to the power of one-half minus π₯ to the power of negative four with respect to π₯. Next, we can simplify this integral slightly. Remember, the integral of a difference between two functions is equal to the difference between their integrals. So we can just integrate each term separately. And, as we said before, we can integrate both of these terms by using the power rule for integration.

We recall the power rule for integration tells us, for constants π and π, where π is not equal to negative one, the integral of ππ₯ to the πth power with respect to π₯ is equal to π times π₯ to the power of π plus one divided by π plus one plus the constant of integration π. We add one to our exponent and then divide by this new exponent. Applying the power rule of integration to the first term in our integrand, we get eight times π₯ to the power of one-half plus one divided by one-half plus one. We then need to subtract the integral of π₯ to the power of negative four with respect to π₯.

We did this by using the power rule for integration. We got π₯ to the power of negative four plus one divided by negative four plus one. And we then combined both of our constants of integration into one constant of integration we called π. And we can simplify this expression. One-half plus one is equal to three over two, and negative four plus one is equal to negative three. So this gives us eight π₯ to the power of three over two divided by three over two minus π₯ to the power of negative three divided by negative three plus π. And we could leave our answer like this. However, we can simplify our expression slightly.

First, instead of dividing by three over two, weβre going to multiply by the reciprocal of three over two. Next, by using our laws of exponents, instead of having π₯ to the power of negative three in our numerator, weβll divide by π₯ cubed. And the reciprocal of three over two is two-thirds. This gives us eight times two-thirds times π₯ to the power of three over two minus one divided by negative three π₯ cubed plus π. And we can simplify this again. Eight times two-thirds is equal to sixteen-thirds.

Next, we can cancel our shared factor of negative one in the numerator and the denominator of our second term. And the last thing weβll do is rewrite π₯ to the power of three over two by using our laws of exponents. Weβll write this as the square root of π₯ cubed. And this gives us our final answer. Weβve shown the integral of eight root π₯ minus one divided by π₯ to the fourth power with respect to π₯ is equal to 16 over three times the square root of π₯ cubed plus one over three π₯ cubed plus π.