Question

Find :
$\int \frac{1}{a^x{b}^x}dx$

Answer 1

Consider the given integral.

$I=\int \frac{1}{a^x{b}^x}dx$ …….. (1)

Let $t=\frac{1}{a^x{b}^x}$

$t=a^{-x}{b}^{-x}$

$\frac{dt}{dx}=a^{-x}(-b^{-x}{\log }_{e}b)+b^{-x}(-a^{-x}{\log }_{e}a)$

$\frac{dt}{dx}=-(\frac{{\log }_{e}b}{a^x{b}^x}+\frac{{\log }_{e}a}{a^x{b}^x})$

$-\frac{a^x{b}^{x}dt}{{\log }_{e}b+{\log }_{e}a}=dx$

$-\frac{dt}{({\log }_{e}b+{\log }_{e}a)t}=dx$

Therefore,

$I=-\frac{1}{({\log }_{e}b+{\log }_{e}a)}\int 1dt$

$I=-\frac{1}{({\log }_{e}b+{\log }_{e}a)}\left(t\right)+C$

On putting the value of $t$, we get

$I=-\frac{1}{({\log }_{e}b+{\log }_{e}a)}\left(\frac{1}{a^x{b}^x}\right)+C$

Hence, this is the answer.