$\int 2^x \ln(x)\, \mathrm{d}x$ 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? 
 A: First you did a mistake here:
$$\int 2^x \ln(x)\, \mathrm{d}x = \frac{\ln(x)2^x}{\ln 2} - \left[ \frac{2^x \ln x}{\ln 2} - \int \ln(x) 2^x\, \mathrm{d}x \right]\Rightarrow \frac{\ln(x)2^x}{\ln 2} -\frac{2^x \ln x}{\ln 2}= 0$$
You can't just cancel the integrals, as you will lose the constant of integration. For example
$$\int \frac{1}{x}\, \mathrm{d}x = \int x^{\prime}\frac{1}{x}\, \mathrm{d}x=
x\frac{1}{x} - \int x\frac{-1}{x^2}\, \mathrm{d}x =1+\int \frac{1}{x}\, \mathrm{d}x$$ 
If you cancel the integrals then $1=0$ which is impossible. When canceling integrals one must never forget the constant of integration $c$. 
In our case $c=0$. To see this,
$$\int 2^x \ln(x)\, \mathrm{d}x = \frac{\ln(x)2^x}{\ln 2} - \frac{2^x \ln x}{\ln 2} + \int \ln(x) 2^x\, \mathrm{d}x \Rightarrow 0=\frac{\ln(x)2^x}{\ln 2} - \frac{2^x \ln x}{\ln 2}+c=c$$
which leads to $0=0$. Why did this come up? You integrated by parts once and then did the reverse and got back to your starting point. Now how can $\int 2^x \ln(x)\, \mathrm{d}x $ be evaluated? It can't be written as a combination of elementary functions (polynomial,exponential,logarithmic,trigonometric and hyperbolic functions and their inverses). I will show this for $\int e^x \ln(x)\, \mathrm{d}x $.
$$\int e^x \ln(x)\, \mathrm{d}x = \int (e^x)^{\prime} \ln(x)\, \mathrm{d}x=e^x\ln x-
\int e^x (\ln(x))^{\prime}\, \mathrm{d}x=e^x\ln x-\int \frac{e^x}{x}\, \mathrm{d}x
$$
The last integral is not elementary as shown by Risch Algorithm. For more information look here: Exponential Integral. And no, I don't think there is any book covering this topic
