Solve ∫x5√1+x2dx
I=∫x5√1+x2dx
Let,
1+x2=x
x4=(u−1)2
2xdx=du
C=12∫(u−1)2√udu
=12∫u2+1−2u√udu
=12∫(u32+u−12−2u12)du
=12(u5252+2u12−2u3232)+c
=15u52+u12−2u323+c
=u1215(3u2+15−10u)+c
=√1+x215[3(1+x4+2x2)+15−10(1+x2)]+c
=√1+x215(3+3x4+6x2+15−10−10x2)+c
=√1+x215(3x4−4x2+8)+c
(B) option.