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Strangely, Mathematica cannot do this definite integral:

Integrate[x/(x^2 + L^2)^(3/2), {x, 0, a}],

while for the indefinite one:

Integrate[x/(x^2 + L^2)^(3/2), x]

the software easily finds the solution:

-(1/Sqrt[L^2+x^2])

Why?

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Probably that's because Mathematica doesn't know what a is, it may be a complex number for example.

If you change your code to:

Integrate[x/(x^2 + L^2)^(3/2), {x, 0, a},Assumptions -> a \[Element] Reals]

you get a solution:

ConditionalExpression[(-1 + Sqrt[1 + a^2/L^2])/(
 Sqrt[1 + a^2/L^2] Sqrt[L^2]), 
 a >= 0 && (Re[L^2/a^2] >= 0 || Re[L^2/a^2] <= -1 || 
    L^2/a^2 \[NotElement] Reals) && Re[L^2] > 0]
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  • $\begingroup$ Thank you very much. And if a and L are real and both >0? Can I impose these assumptions? $\endgroup$ – umby May 23 '18 at 16:39
  • $\begingroup$ sure you can: Integrate[x/(x^2 + L^2)^(3/2), {x, 0, a}, Assumptions -> {a > 0, L > 0}] $\endgroup$ – Fraccalo May 23 '18 at 16:40

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