I found this problem by a typo. My homework problem was $\int 2^x \ln(2) \, \mathrm{d}x$ which is $2^x + C$ by the Fundamental Thm of Calculus. I want to be able to solve what I wrote down incorrectly in my homework.
What I wrote for my homework is $\int 2^x \ln(x)\, \mathrm{d}x$ and What I Want to solve, plus I got it wrong. :(
I used integration by parts. $$\int u \, \mathrm{d}v = uv - \int v\, \mathrm{d}u$$
$$\begin{array}{l l} u = \ln(x) & du = \frac{1}{x}\mathrm{d}x \\ \mathrm{d}v = 2^x\mathrm{d}x & v = \frac{2^X}{\ln (2)} \\ \end{array}$$
I got this integral:
$$\frac{\ln(x)2^x}{\ln 2} - \int \frac{2^x}{x\ln 2}\, \mathrm{d}x$$
Another round of integration of parts:
$$\begin{array}{l l} u = \frac{2^x}{\ln 2} & du = 2^x\mathrm{d}x \\ \mathrm{d}v = \frac{1}{x}\mathrm{d}x & v = \ln(x) \end{array} $$
$$\int 2^x \ln(x)\, \mathrm{d}x = \frac{\ln(x)2^x}{\ln 2} - \left[ \frac{2^x \ln x}{\ln 2} - \int \ln(x) 2^x\, \mathrm{d}x \right]$$
My final answer is
$$ \frac{\ln(x)2^x}{\ln 2} -\frac{2^x \ln x}{\ln 2}= 0$$
What did I do wrong?