Answer
[tex]\int (\frac{x^3}{4}+\frac{2x^2}{3}-1)dx[/tex]Take the integral of each
[tex]\int \frac{x^3}{4}dx+\int \frac{2x^2}{3}dx-\int 1dx[/tex][tex]\int \frac{x^3}{4}dx=\frac{x^4}{16}[/tex][tex]\int \frac{2x^2}{3}dx=\frac{2x^3}{9}[/tex][tex]\int 1dx=x[/tex]Now properly combine them back
[tex]\frac{x^4}{16}+\frac{2x^3}{9}-x+c[/tex]The final answer
Option D