Hi Selena, this integral is a little tricky but here goes:
Make the substitution u = x + 1 du = dx We'll also need that x = u - 1
∫(u-1)2/u du now expand the numerator
∫(u2 - 2u + 1)/u du now break up into 3 terms
∫(u2/u -2u/u + 1/u)du
∫(u -2 + 1/u)du = 1/2u2 - 2u + ln|u| + C now sub back to x
= 1/2(x + 1)2 - 2(x + 1) + ln|x + 1| + C now expand and combine like terms
= 1/2(x2 + 2x + 1) -2x -2 + ln|x+1| + C
= 1/2x2 + x + 1/2 -2x -2 + ln|x+1| + C
= 1/2x2 - x - 3/2 + ln|x+1| + C
which actually look like one of the choices! What a relief.
The real trick here that is kind of different is solving u=x+1 for x so we can substitute the x2 term. Integration is a big bag of tricks and you have to work a lot of them to know what to do. Good Luck!