Leading-order equation

[Amine A]

Hello,
Anyone can understand the passage from line 16 to line 17 please ?

 

[prex]

Hint: the indefinite integral of (1+kx)/(1+hx) is (kx+(1-k/h)log(1+kx))/h+C

prex
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[btrueblood]

Yeah, but the left side of the equation stayed at d^2, should it not become the first derivative after the integration?

 

[onatirec]

(d/dz)^2(T) =/= (dT/dz)^2

 

[prex]

Take the integral in ∂z of both sides, then multiply both integrands by ∂T/∂z , so that the right integral is in ∂T, and you can calculate it as above, and on the left you have (∂T/∂z)[sup]2[/sup] I cannot explain the factor 2 in the result, though.

prex
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