L=∫1+(y′)2dxy=2xy′=2x1L=∫0211+2x1dx
This can also be written as L=∫0212x1+1dx
we apply linearity
21∫x1+2dx
let u = x1+2 , dx = -2x1+2x2du,dx=−2∫(u2−2)u2du
we integrate ∫(u2−2)u2du=∫((u2−2)2u2−2+(u2−2)22)du=
∫((u2−2)1+(u2−2)22)du
we split and integrate differently.
∫((u2−2)1du⟹223Ln(u−2)−223Ln(u+2)
also
∫(u2−2)22du⟹∫((u2−2)1du⟹223Ln(u−2)−223Ln(u−2)−4(u+2)1−4(u−2)1
but substituting the solution of the integrations back, we get
223Ln(u−2)−223Ln(u+2)+223Ln(u−2)−223Ln(u−2)−4(u+2)1−4(u−2)1
recall, u= x1+2 then we substitute u in terms of x back and introduce the lower and upper limits
223Ln(u−2)−223Ln(u+2)+223Ln(u−2)−223Ln(u−2)−4(u+2)1−4(u−2)1∣021
We have.
21+4Ln(2+1)−Ln(2−1)
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