Answer
$$\int\log_2xdx=x\log_2x-\frac{x}{\ln2}+C$$
Work Step by Step
$$A=\int\log_2xdx$$
Take $y=f^{-1}(x)=\log_2x$
So $x=f(y)=2^y$
Applying the formula $\int f^{-1}(x)dx=xf^{-1}x-\int f(y)dy$, we have
$$A=x\log_2x-\int2^ydy$$ $$A=x\log_2x-\frac{2^y}{\ln2}+C$$ $$A=x\log_2x-\frac{2^{\log_2x}}{\ln2}+C$$ $$A=x\log_2x-\frac{x}{\ln2}+C$$
