 In this video, we provide the solution to question number nine for practice exam number one for math 1220. We have to evaluate the indefinite integral of 3 minus 2x over x squared minus 3x to the one-third power with respect to x. I'm going to do this using a U substitution. My thought process is I have this x squared minus 3x in the denominator. It's raised to the power of one-third. It'd be really nice and hunky-dory if its derivative is on top. I'm going to try U substitution. That would be my dream come true right now. U is x squared minus 3x. Then du would equal 2x minus 3 dx. As I come over here and analyze this, that's almost what I have. The numerator is 3 minus 2x. That's the function I'm looking for. Just times it by negative one. What that does for us is then we can rewrite this integral here. If I take out the negative sign, you get negative 2x minus 3 dx sitting on top of the x squared minus 3x to the one-third power. That's then perfect for what we need. We're going to get negative the integral. The top then became that negative du. I already brought the negative sign out. The denominator is u to the one-third power. Instead, I think of this as negative the integral of u to the negative one-third power du. I can then apply the power rule to the u for which we're going to increase the power by one. We take u negative one-third plus one. That's going to give us a positive two-thirds. We then divide by two-thirds. Don't forget your arbitrary constant now. Then rewriting that if you divide by two-thirds, that's the same thing as times in by three-halves. We get negative three-halves times u to the two-thirds plus a constant. Now the anti-derivative is computed. We just need to plug back in the original value for u. We end up with an anti-derivative of negative three-halves times x squared minus 3x raised to the two-thirds power plus an arbitrary constant. Do remember the constant there. It's not full credit if you don't have that plus c.