# Thread: Integral of arbitary powers

1. ## Integral of arbitary powers

the integral b=1, a=0 of (1/2)[(a^x-a^-x)/(a^x+a^-x)]

2. Originally Posted by akhayoon
the integral b=1, a=0 of (1/2)[(a^x-a^-x)/(a^x+a^-x)]
do you mean $\int_0^1 \frac {a^x - a^{-x}}{a^x + a^{-x}}~dx$ ?

using a for the limits and then a again for the constants in the integral is confusing. be specific and clear

if that is your integral, use the substitution $u = a^x + a^{-x}$

3. after the substitution will the new bounds be

$lna~ \int_2^\frac {a^2+1} {a} \frac {du} {u}$ ?

$lna~ [ln|u|]$ ? with those bounds I mentioned of course

4. Originally Posted by akhayoon
after the substitution will the new bounds be

$lna~ \int_2^\frac {a^2+1} {a} \frac {du} {u}$ ?

$lna~ [ln|u|]$ ? with those bounds I mentioned of course