# Integrate and InverseFunction

I solve an easy example to check that Mathematica can understand the difference between ArcTan and ArcTanh but as you see here I can't get Arctanh??!! Even I don't know why the inverse function is not working! thanks for helping

g = Integrate[1/((-(3/2))*x^2 + (3/2)*a*x+ b/2), x, Assumptions -> {a > 0, b > 0}]
Assuming[{a > 0, b >0},InverseFunction[g][x]]


a,b are positive.

• g isn't a function; it's a symbolic expression. Try g[x_] = Integrate[...] instead (after Clear[g]); Nov 18, 2021 at 14:42
• My problem is the answer is Tanh[] but it gives me Tan[] !!! tan(ix) = i tanh x Nov 18, 2021 at 15:12

You'll get the result in terms of the hyperbolic functions if you use FullSimplify:

Clear[g]
g[x_] = Assuming[
a > 0 && b > 0,
FullSimplify@Integrate[1/((-(3/2))*x^2 + (3/2)*a*x + b/2), x]
]
Assuming[
a > 0 && b > 0,
FullSimplify[InverseFunction[g][x]]
] // Normal


-((4 ArcTanh[(a - 2 x)/Sqrt[a^2 + (4 b)/3]])/Sqrt[9 a^2 + 12 b])

1/6 (3 a + Sqrt[9 a^2 + 12 b] Tanh[1/4 Sqrt[9 a^2 + 12 b] x])

• Thank you, so assuming will work here Nov 19, 2021 at 15:51