f(x)(1+x2)=1+x∫0f2(t)1+t2 dt
Differentiate w.r.t. x
(1+x2)⋅(f′(x))−2xf(x)(1+x2)2=f2(x)1+x2
dydx−(2x1+x2)y=y2
Let −1y=t
dtdx+(2x1+x2)t=1
Solution of the above equation is :
−1y(1+x2)=x33+x+C
Since f(0)=1, we get C=−1
Hence, f(x)=−3(1+x2)x3+3x−3
Equation of tangent at (1,−6) to y=f(x) is y=30x−36
△=1085 sq.units