i tried to figure it out but i can't. i need to show the work and i don't know how to do it.​

Integration Formula:[tex]\int\limits {x}^{n} \, dx =\frac{x^{n+1}}{n+1} + C[/tex]Integrate each term:[tex]\int\ {x^{2}} \, dx = \frac{x^{3}}{3}[/tex][tex]\int\ {x} \, dx =\frac{x^2}{2}[/tex][tex]\int\ {5} \, dx = 5x[/tex][tex]\int\ {x} \, dx =\frac{x^2}{2}[/tex][tex]\int\ {2} \, dx = 2x[/tex]Altogether this becomes:[tex]\frac{\frac{x^3}{3} - \frac{x^2}{2}-5x }{ \frac{x^2}{2}+2x}[/tex]Your limits are 1 and 2. Substitute both numbers into the equation and minus them from each other.x = 2:[tex]\frac{\frac{2^3}{3} - \frac{2^2}{2}-5(2) }{ \frac{2^2}{2}+2(2)}[/tex]-x = 1:[tex][tex]\frac{\frac{1}{3} - \frac{1}{2}-5 }{ \frac{1}{2}+2}[/tex][/tex]x = 2:[tex]\frac{\frac{8}{3} - \frac{4}{2}-10 }{ \frac{4}{2}+4}[/tex]-x = 1: [tex]\frac{\frac{8}{3} - 2-10 }{ 2+4}[/tex]=-1.212317928Now convert all the answers to integers and see which one matches -1.212317928:Because it's a negative number, we know it can only be A or B.(A) = -1.212317928(B) = -1.19047619The answer is A: [tex]-\frac{3}{2} +In\frac{4}{3}[/tex](A) - 3/2 = ln 4/3