L=β«1+(yβ²)2βdxy=2xβyβ²=2xβ1βL=β«021ββ1+2x1ββdx
This can also be written as L=β«021ββ2x1β+1βdx
we apply linearity
2β1ββ«x1β+2βdx
let u = x1β+2β , dx = -2x1β+2x2βdu,dx=β2β«(u2β2)u2βdu
we integrate β«(u2β2)u2βdu=β«((u2β2)2u2β2β+(u2β2)22β)du=
β«((u2β2)1β+(u2β2)22β)du
we split and integrate differently.
β«((u2β2)1βduβΉ223βLn(uβ2β)ββ223βLn(u+2β)β
also
β«(u2β2)22βduβΉβ«((u2β2)1βduβΉ223βLn(uβ2β)ββ223βLn(uβ2β)ββ4(u+2β)1ββ4(uβ2β)1β
but substituting the solution of the integrations back, we get
223βLn(uβ2β)ββ223βLn(u+2β)β+223βLn(uβ2β)ββ223βLn(uβ2β)ββ4(u+2β)1ββ4(uβ2β)1β
recall, u= x1β+2β then we substitute u in terms of x back and introduce the lower and upper limits
223βLn(uβ2β)ββ223βLn(u+2β)β+223βLn(uβ2β)ββ223βLn(uβ2β)ββ4(u+2β)1ββ4(uβ2β)1ββ£021ββ
We have.
2β1β+4Ln(2β+1)βLn(2ββ1)β
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