Let g be a differentiable function satisfies ∫0x (x−t+1)g(t)dt=x4+x2∀x≥0,  then the value of ∫01 12g1(x)+g(x)+10dx=

x∫0x g(t)dt+∫0x (1−t)g(t)dt=x4+x2 differentiate with respect to x∫0x g(t)dt+xg⁡(x)+1-xgx=4x3+2x again differentiate with respective to x⇒g′(x)+g(x)=12x2+2∴∫01 12g1(x)+g(x)+10dx=∫01 1212x2+1dx=tan-1x01=π4