The answer is where all Math begins

$The\quad integrand\quad can\quad be\quad simplified\quad as\quad follows-\\ \Rightarrow \frac { \left( 32\sqrt { 7 } \left( x^{ 3 }+x^{ 4 } \right) +8x^{ 2 }\left( \sqrt { 31 } -17\sqrt { 7 } \right) +6x\left( \sqrt { 31 } -12\sqrt { 7 } \right) +\sqrt { 31 } +164\sqrt { 7 } \right) }{ 2\left( 1+2x \right) \left( 41-18x-34x^{ 2 }+8\left( x^{ 3 }+x^{ 4 } \right) \right) } \\ \\ \Rightarrow \frac { \left( 32\sqrt { 7 } x^{ 3 }+32\sqrt { 7 } x^{ 4 }-136\sqrt { 7 } x^{ 2 }-72\sqrt { 7 } x+164\sqrt { 7 } \right) }{ 32x^{ 5 }+48x^{ 4 }-120x^{ 3 }-140x^{ 2 }+128x+82 } +\frac { \left( 8\sqrt { 31 } x^{ 2 }+6\sqrt { 31 } x+\sqrt { 31 } \right) }{ 32x^{ 5 }+48x^{ 4 }-120x^{ 3 }-140x^{ 2 }+128x+82 } \\ \\ \Rightarrow \sqrt { 7 } \frac { \left( 32x^{ 3 }+32x^{ 4 }-136x^{ 2 }-72x+164 \right) }{ 32x^{ 5 }+48x^{ 4 }-120x^{ 3 }-140x^{ 2 }+128x+82 } +\sqrt { 31 } \frac { \left( 8x^{ 2 }+6x+1 \right) }{ 32x^{ 5 }+48x^{ 4 }-120x^{ 3 }-140x^{ 2 }+128x+82 } \\ \\ Which\quad after\quad a\quad tedious\quad long\quad divison\quad simplifies\quad to-\\ \Rightarrow \frac { 2\sqrt { 7 } }{ \left( 2x+1 \right) } +\frac { \sqrt { 31 } \left( 4x+1 \right) }{ 16x^{ 4 }+16x^{ 3 }-68x^{ 2 }-36x+82 } \\ \\ Therefore\quad the\quad problem\quad simplifies\quad to-\\ \\ \Rightarrow \int { \frac { 2\sqrt { 7 } }{ \left( 2x+1 \right) } +\frac { \sqrt { 31 } \left( 4x+1 \right) }{ 16x^{ 4 }+16x^{ 3 }-68x^{ 2 }-36x+82 } } dx\\ \\ \Rightarrow \int { \frac { 2\sqrt { 7 } }{ \left( 2x+1 \right) } } dx\quad +\int { \frac { \sqrt { 31 } \left( 4x+1 \right) }{ 16x^{ 4 }+16x^{ 3 }-68x^{ 2 }-36x+82 } } dx\\ The\quad first\quad integral\quad is\quad the\quad derivative\quad of\quad \sqrt { 7 } \ln { (2x+1) } \quad and\quad the\quad second\quad integral\quad further\quad simplifies\quad as\quad follows..\\ \\ \Rightarrow \int { \frac { \sqrt { 31 } \left( 4x+1 \right) }{ 16x^{ 4 }+16x^{ 3 }-68x^{ 2 }-36x+82 } } dx\quad \\ \Rightarrow \frac { \sqrt { 31 } }{ 2 } \int { \frac { 2\left( 4x+1 \right) }{ (16x^{ 4 }+16x^{ 3 }-68x^{ 2 }-36x+81)+1 } } dx\\ \Rightarrow \frac { \sqrt { 31 } }{ 2 } \int { \frac { \left( 8x+2 \right) }{ 1+\left( 4x^{ 2 }+2x-9 \right) ^{ 2 } } } dx\\ This\quad is\quad just\quad the\quad derivative\quad of\quad \frac { \sqrt { 31 } }{ 2 } \arctan { (4x^{ 2 }+2x-9) } \\ \\ Therefore\quad the\quad integral\quad finally\quad is\quad \left[ \sqrt { 7 } \ln { (2x+1) } \quad +\quad \frac { \sqrt { 31 } }{ 2 } \arctan { (4x^{ 2 }+2x-9) } \right] \quad which\quad on\quad putting\quad proper\quad limits\quad and\quad evaluating\quad equals-\\ \frac { \sqrt { 31 } \pi }{ 4 } +\sqrt { 7 } \ln { \left( \frac { \left( \sqrt { 41 } +1 \right) }{ \sqrt { 33 } +1 } \right) } \\ \therefore \boxed { A=31\\ B=4\\ C=7\\ D=41\\ E=33 } \quad And\quad thus,\quad \boxed { \quad \frac { (A-8B) }{ C-D+E } =1\quad } \\ \\ \\$