The correct option is C π2√2−165√2
I=∫√2−√22x7+3x6−10x5−7x3−12x2+x+1x2+2dx
=∫√2−√22x7−10x5−7x3−12x2+x+1x2+2dx+∫√2−√23x6−12x2+1x2+2dx
=0+2∫√203x2−12x2+1x2+2dx=2∫√203x2(x4−4)+1x2+2dx
=2∫√20(3x2(x2−2)+1x2+2)dx=2∫√20(3x4−6x2+1x2+2)dx
=2(3x55−2x3+1√2tan−1(x2))√20=π2√2−16√25