∫ 3 x 1 - 9 x d x =

Let $I=\int \frac{3x}{\sqrt{1-9^x}}dx$

$=\int \frac{3x}{\sqrt{1-(3^x{)}^2}}dx$

Put $3^x=t$, we get

$3^x \log 3 dx=dt$

$3^{x}dx=\frac{1}{\log 3}dt$

$3^{x}dx=(\log 3{)}^{-1} dt$

$I=\int \frac{(\log 3{)}^{-1}}{\sqrt{1-t^2}}dt$

$=\int \frac{1}{\sqrt{a^2-x^2}}dx= {\sin }^{-1}\left(\frac{x}{a}\right)+c)$

$I=(\log 3{)}^{-1} Si{n}^{-1}(t)+c$

Again put $t=3^x$, we get

$I=(\log 3{)}^{-1} Si{n}^{-1}(3^x)+c$.