Answer
$\dfrac{7}{3}+\dfrac{e^2-e}{2}$
Work Step by Step
$W=\int_C F\cdot dr=\int_0^{1} (y^2+1)^2(2y dy)+y[e^{y^2+1}] dy$
or, $=\int_0^{1} 2y \times (y^2+1)^2+y \times [e^{y^2+1}] dy$
Suppose, $y^2+1=t; 2y dy =dt$
$=\int_1^2 t^2+\dfrac{e^t}{2} dt$
$=[\dfrac{t^3}{3}+\dfrac{e^t}{2}]_1^{2}$
$=[\dfrac{2^3}{3}+\dfrac{e^2}{2}]-[\dfrac{1^3}{3}+\dfrac{e^1}{2}]$
$=\dfrac{7}{3}+\dfrac{e^2-e}{2}$
