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Hello: Is there a way Mathematica to transform a definite integral in t into a function in x, per the FTC1 equation:

g[x_] = (
\*SubsuperscriptBox[\(\[Integral]\), \(2\), \(x\)]\(
\*FractionBox[\(
\*SuperscriptBox[\(t\), \(2\)] - 1\), \(
\*SuperscriptBox[\(t\), \(2\)] + 1\)] \[DifferentialD]t\)\)

The software just spins when I try this, either as '=' or ':='. I've also tried both:

g[x] = Integrate[f[t], {t, 2, x}]
g[x] = Integrate[f[t], {x, 2, x}]

The first just spins. The second returns:

((-1 + t^2) (-2 + x))/(1 + t^2)

...which is interesting, but not the function completely in x I was looking for.

I guess another way of putting my question is, can we get M. to handle upper and/or lower limits of integration as variables?

Thanks, TB

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    $\begingroup$ Perhaps Integrate[(t^2 - 1)/(t^2 + 1), {t, 2, x}, GenerateConditions -> False]? $\endgroup$
    – Carl Woll
    Apr 18 at 19:25
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    $\begingroup$ You have to define a function with Blank, g[x_] not g[x]. $\endgroup$
    – Akku14
    Apr 18 at 19:48
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    $\begingroup$ Don't know what you exactly want. Maybe this g[x_] = Integrate[f'[t], {t, 2, x}] yields -f[2] + f[x] $\endgroup$
    – Akku14
    Apr 18 at 20:02
  • $\begingroup$ Thank you, all Yes, 'GenerateConditions' does the trick. $\endgroup$
    – Tom Barson
    Apr 18 at 23:14

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