Solving $\displaystyle\int \dfrac{1+x \cdot \ln{x}}{x \cdot \ln{x}} dx$ with the $ \displaystyle{\int} \dfrac{f^{\prime} }{f} dx= \ln{f} $ identitiy.

Question

I know how to solve this integral with u-substitution but im pretty sure it can be solved using the fact that if $ f(x) = x \cdot \ln{x} $ then $f^{\prime} = \ln{x} +1 $. However i don't know how.

Answer 1

Write

$$\frac{1+x\ln x}{x\ln x}=1+\frac{1}{x\ln x}=1+\frac{1/x}{\ln x}$$

Answer 2

Add and subtract a log in the numerator like so:

$$\int \frac{1+\ln x - \ln x + x\ln x}{x\ln x}\:dx = \int \frac{1+\ln x}{x\ln x}+1-\frac{1}{x}\:dx = \ln(x\ln x)+x-\ln x + C$$ $$ = \ln \ln x + x + C$$

Answer 3

You can just use the following:$$I=\int\frac{1+x\ln x}{x\ln x}dx=\int\frac{1/x}{\ln x}dx+x=\int \frac{d(\ln x)}{\ln x}dx+x=\ln(\ln x)+x+C.$$