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I need a rusult with the following form

D[Integrate[f[t], {t, low[x], upper[x]}], x]
(*Out[]=-f[low[x]] (low^\[Prime])[x]+f[upper[x]] (upper^\[Prime])[x]*)

My codes are

Clear["Global`*"]
(*a=110.;b=55.;d=1.;m1=18.;m2=42.;m=m2/m1;
*)
w[l_, e_] := (-((m1*a)/2) Log[1 - (l^(-4) + 2*l^2 - 3)/a] - (m2*b)/
      2 Log[1 - (l^-4*e^4 + 2 l^2*e^-2 - 3)/b])/m1
dw[l_, e_] := D[w[l, e], l]
f[l_, e_] := dw[l, e]/(1 - l^3)
sup[x_] := ((d + x^3)/(1 + d))^(1/3)
intf[x_, e_] := Integrate[f[l, e], {l, x, sup[x]},
                Assumptions-> a > 0 && b > 0 && d > 0 && m1 > 0 && m2 > 0]
D[intf[x, e], x]

How could I get the desired result for D[intf[x, e], x]? Any suggestiones would be much appricated!

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Perhaps you can inactivate the integral:

intf[x_, e_] := Inactive[Integrate][f[l, e], {l, x, sup[x]}]

D[intf[x, e], x] //TeXForm

$\frac{x^2 \left(\frac{\text{m1} \left(4 \sqrt[3]{\frac{d+x^3}{d+1}}-\frac{4}{\left(\frac{d+x^3}{d+1}\right)^{5/3}}\right)}{2 \left(1-\frac{2 \left(\frac{d+x^3}{d+1}\right)^{2/3}+\frac{1}{\left(\frac{d+x^3}{d+1}\right)^{4/3}}-3 }{a}\right)}+\frac{\text{m2} \left(\frac{4 \sqrt[3]{\frac{d+x^3}{d+1}}}{e^2}-\frac{4 e^4}{\left(\frac{d+x^3}{d+1}\right)^{5/3}}\right)}{2 \left(1-\frac{\frac{e^4}{\left(\frac{d+x^3}{d+1}\right)^{4/3}}+\frac{2 \left(\frac{d+x^3}{d+1}\right)^{2/3}}{e^2}-3}{b}\right)}\right)}{(d+1) \text{m1} \left(\frac{d+x^3}{d+1}\right)^{2/3} \left(1-\frac{d+x^3}{d+1}\right)}-\frac{\frac{\text{m1} \left(4 x-\frac{4}{x^5}\right)}{2 \left(1-\frac{\frac{1}{x^4}+2 x^2-3}{a}\right)}+\frac{\text{m2} \left(\frac{4 x}{e^2}-\frac{4 e^4}{x^5}\right)}{2 \left(1-\frac{\frac{e^4}{x^4}+\frac{2 x^2}{e^2}-3}{b}\right)}}{\text{m1} \left(1-x^3\right)}$

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  • $\begingroup$ Thank you, It really works! $\endgroup$ – keanhy14 Jun 17 at 6:35

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