Integrate $\int \frac{dx}{3^x-8}$

Question

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?

Answer 1

Let $3^x-8=u\implies 3^x\ln(3)\ dx=du$ or $dx=\frac{1}{(u+8)\ln (3)}\ du$ $$\int \frac{1}{3^x-8}\ dx=\int \frac{1}{u(u+8)\ln 3}\ du $$ $$=\frac{1}{\ln 3}\int \frac{1}{u(u+8)}\ du $$ $$=\frac{1}{8\ln 3}\int\left( \frac{1}{u}-\frac{1}{u+8}\right)\ du $$ $$=\frac{1}{8\ln 3}\ln\left| \frac{u}{u+8}\right| +C$$ $$=\frac{1}{8\ln 3}\ln\left| \frac{3^x-8}{3^x}\right| +C$$

Answer 2

As a way of getting around your error, use the fact that $$\frac 1{3^x-8}=\frac{3^{-x}}{1-8\cdot 3^{-x}}$$

then let $u=1-8\cdot3^{-x}$.

Answer 3

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

$$ \int \frac{1}{3^x - 8} = -\frac{1}{8} \left( x + \int \frac{3^x dx}{8 - 3^x} \right) = -\frac{x}{8} +\frac{1}{8 \ln 3} \int \frac{d(8-3^x)}{8 - 3^x} = -\frac{x}{8} + \frac{\ln (8-3^x)}{8 \ln 3} + C $$