<s>[INST] Integrate $\int \frac{dx}{3^x-8}$ I have a problem with solving this Integrate
$$\int \frac{dx}{3^x-8}$$
At first I use substitute $$3^x =t $$
$$\log_3{t} = x$$
$$\frac{1}{t\ln{3}}dt = dx$$
Next:
$$\int \frac{dt}{t^2\ln3} - \int\frac18dx$$
$$=\frac{1}{\ln3}\int \frac{dt}{t^2} - \int\frac18dx$$
$$=\frac{1}{\ln3} \left(\frac{-1}{t}\right) - \frac18$$
$$=\frac{1}{\ln3} \left(\frac{-1}{3^x}\right) - \frac18$$
But Wolfram alpha gives me different result. Where did I go wrong? 
 [/INST] Notice $\frac{1}{3^x - 8} = -\frac{1}{8 - 3^x} = - \frac{1}{8} \frac{8 - \mathbf{3^x} + \mathbf{3^x} }{(8-3^x)}  = -\frac{1}{8} \left( 1 + \frac{3^x}{8-3^x} \right)$ and so 
$$ \int \frac{1}{3^x - 8} = -\frac{1}{8} \left( x + \int \frac{3^x dx}{8 - 3^x} \right) = -\frac{x}{8} +\frac{1}{8 \ln 3} \int \frac{d(8-3^x)}{8 - 3^x} = -\frac{x}{8} + \frac{\ln (8-3^x)}{8 \ln 3} + C $$
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