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

Determine β« 7β(π₯Β³) dπ₯.

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### Video Transcript

Determine the integral of seven times the square root of π₯ cubed with respect to π₯.

The question wants us to integrate seven multiplied by the square root of π₯ cubed with respect to π₯, but we donβt know how to integrate this directly. This means weβre going to need to manipulate or integrand into a form which we can integrate. Luckily, we can do this by using our laws of exponents. First, we recall that the square root of π is the same as saying π to the power of one-half. Using this, we can rewrite our integral as the integral of seven times π₯ cubed to the power of one-half with respect to π₯.

But weβre still not done. We still canβt integrate this directly. We need to use another one of our laws of exponents. We recall that π to the power of π all raised to the power of π is equal to π to the power of π times π. Using this, we can show that π₯ cubed all raised to the power of one-half is equal to π₯ to the power of three over two. So we now need to integrate seven times π₯ to the power of three over two with respect to π₯. And we can do this 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 a constant of integration π. We add one to our exponent of π₯ and then divide through by this new exponent of π₯. In our case, the exponent of π₯ is three over two, so weβll set π to be three over two. So applying our power rule of integration with π equal to seven and π equal to three over two, we get seven times π₯ to the power of three over two plus one divided by three over two plus one plus a constant of integration π.

And we can simplify this since three over two plus one is equal to five over two. So this gives us seven times π₯ to the power of five over two divided by five over two plus π. And to make this easier, instead of dividing by the fraction five over two, weβll multiply by the reciprocal. And the reciprocal of five over two is two-fifths. This gives us seven times two-fifths times π₯ to the power of five over two plus π. And, finally, we can simplify seven times two-fifths to be 14 over five. Therefore, weβve shown the integral of seven times the square root of π₯ cubed with respect to π₯ is equal to 14π₯ to the power of five over two divided by five plus the constant of integration π.