Factor polynomial then partial fractions

7 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = a x ^ { 3 } - x ^ { 2 } + 4 x - a$$ where \(a\) is a constant. It is given that \(( 2 x - 1 )\) is a factor of \(\mathrm { p } ( x )\).

1. Find the value of \(a\) and hence factorise \(\mathrm { p } ( x )\).
2. When \(a\) has the value found in part (i), express \(\frac { 8 x - 13 } { \mathrm { p } ( x ) }\) in partial fractions.

7. (i) Show that ( \(2 x + 3\) ) is a factor of ( \(\left. 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 \right)\) and hence, simplify $$\frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } .$$ (ii) Show that $$\int _ { 2 } ^ { 5 } \frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } \mathrm {~d} x = \ln k$$ where \(k\) is an integer.

13 By first factorising completely \(x ^ { 3 } + x ^ { 2 } - 5 x + 3\), find \(\int \frac { 2 x ^ { 2 } + x + 1 } { x ^ { 3 } + x ^ { 2 } - 5 x + 3 } \mathrm {~d} x\).