# Video: APCALC02AB-P1A-Q08-705172353107

Calculate β«(π₯β΄ + 2π₯)Β²dπ₯.

02:04

### Video Transcript

Calculate the indefinite integral of π₯ to the power of four plus two π₯ all squared with respect to π₯.

Weβll start by expanding the brackets inside the integral. This gives us the indefinite integral of π₯ to the power of eight plus four π₯ to the power of five plus four π₯ squared with respect to π₯. This is simply a polynomial. And so, when integrating each term, we simply increase the power by one and divide by the new power. Now, the only case which we have to be careful for when using this rule is when we have π₯ to the power of negative one. However, none of our times have a power of negative one. Therefore, weβre okay to use this rule here.

When integrating π₯ to the power of eight, we increase the power by one to give us π₯ to the power of nine and then divide by the new power. So thatβs dividing by nine. In our next term, we have four π₯ to the power of five. Our constant four will remain. And then, for the π₯ to the power of five, we increase the power by one, giving us π₯ to the power of six. And then, divide by the new power, so thatβs dividing by six. For our final term, the constant four will, again, remain. Increasing the power by one gives us π₯ cubed. And then, we divide by three.

Weβve now integrated every term. However, we mustnβt forget to add on our constant of integration, π, since this is an indefinite integral. Now, we are very nearly at our solution. However, we can do some simplifying and rearranging to make this look a bit better. We can start by cancelling a factor of two in the second term. Next, we can write each of the fractions with a common denominator. And this enables us to combine the fractions into one, giving us π₯ to the power of nine plus six π₯ to the power of six plus 12π₯ cubed all over nine plus π.

For our final step of simplifying, we can factor out a factor of π₯ cubed. We arrive at our solution that the indefinite integral of π₯ to the power of four plus two π₯ all squared, with respect to π₯, is equal to π₯ cubed multiplied by π₯ to the power of six plus six π₯ cubed plus 12 all over nine plus π.