# Can Mathematica return a function in x from a definite integral in t (Fundamental Theorem of Calculus, part 1)

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

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