Find $\int \frac{\ln\left(\frac{{e^x}}{x}\right)}{{x}}dx$

Question

Find $\displaystyle\int \frac{\ln\left(\frac{{e^x}}{x}\right)}{{x}}dx$.

So, here is what I did: \begin{align} I & := \int \frac{\ln(e^x)-\ln(x)}{{x}}dx = \int \frac{x-\ln(x)}{{x}}dx = \int 1 - \frac{\ln(x)}{x}dx \\ & = \int 1 - \int\frac{\ln(x)}{x}dx =x - \int\frac{\ln(x)}{x}dx. \end{align} Now, let $u = \ln(x)$. Then we have $du = 1/x dx$ and $dx = x du$ therefore \begin{align} I = x - \int\frac{u}{x}x\ du = x - \int\frac{u}du = x - \frac{u^2}{2} = x - \frac{\ln^2(x)}{2} = x - \ln(x) \end{align}

However, when I checked on Symbolab it gives me $ x - \frac{\ln^2(x)}{2} - \frac{1}{2} + C$.

I couldn't quite get where I go wrong. If you can explain it to me it would be great. Thanks.

Answer 1

There's only a small mistake: $$ x - \frac{\ln^2(x)}{2} \ne x- \ln(x). $$ As pointed out in the other answer, you were probably thinking $$\ln(x)^2 \overset{(\ddagger)}{=} \ln(x^2) \overset{(\star)}{=} 2 \ln(x).$$ The step $(\star)$ is correct in general, but ($\ddagger$) is not. Up until that step everything is correct.

When evaluating a indefinite integral you have to add the arbitrary constant $+C$, which you forgot. Since $$ x - \frac{\ln^2(x)}{2} - \frac{1}{2} + C = x - \frac{\ln^2(x)}{2} + \tilde{C} $$ for some other constant $\tilde{C}$, the result is correct. Also notice that SymboLab also has a $+C$ in their solution.

Answer 2

$\dfrac{(\ln x)^2}2$ will be $=\dfrac{\ln(x^2)}2$

if $(\ln x)^2=2\ln x\iff \ln x=0,2$

So, in general they are not same

and we know $\ln(x^m)=m\ln x$

Answer 3

How did you get from $\frac{\ln^2{x}}{2}$ to $\ln{x}$? The 2 is on the logarithm: $\ln^2{x}=(\ln{x})^2$. It's not on the $x$. So, you can't bring the $2$ out front like that. Your answer one step back looked fine. Just add a constant of integration to it and it's the same answer as that of Symbolab's:

$$ x-\frac{\ln^2{x}}{2}+K=x-\frac{\ln^2{x}}{2}-\frac{1}{2}+C. $$

Never forget that there always must be a constant of integration at the end of your answer.

Think about what happens if you differentiate both results. $K$ and $-\frac{1}{2}+C$ are both numbers and numbers, as you must know, always differentiate to zero. So, in both cases, you end up differentiating exactly the same function.