Let's now take the definite integral from two to four of six plus x squared over x to the third power dx. At first this might seem pretty daunting. I have this rational expression, but if we just rewrite this, it might jump out at you how this could be a little bit simpler. So this is equal to the integral from two to four of six over x to the third power plus x squared over x to the third power dx. I just separated this numerator out. I just divided each of those terms by x to the third power. And this I could rewrite.