IF=e∫(1+x2)22xdx=e1+x2−1y⋅e1+x2−1=∫x⋅e1+x21⋅e1+x2−1dxy⋅e1+x2−1=2x2+c(0,0)⇒C=0y(x)=2x2e1+x21f(x)=2x2 
A=∫−24(x+4)−2x2dx=18
JEE Main 2024 — Mathematics Calculus
Let y=y(x) be the solution of the differential equation dxdy+(1+x2)22xy=xe(1+x2)1;y(0)=0.
Then the area enclosed by the curve f(x)=y(x)e−(1+x2)1 and the line y−x=4 is__________
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