If we want to integrate

$\LARGE \int \frac{dx}{xlnx}$

We may use the integration by parts technique instead of substitution to get $\LARGE I=\int { \frac { dx }{ x\ln { x } } } =\frac { 1 }{ \ln { x } } \int { \frac { dx }{ x } } -\int { \left( -\frac { 1 }{ { (\ln { x } ) }^{ 2 } } \frac { 1 }{ x } \int { \frac { dx }{ x } } \right) dx }\\ \\ \LARGE=\frac { 1 }{ \ln { x } } \ln { x } +\int { \frac { dx }{ x\ln { x } } }\\ \\ \LARGE=1+I \\ \\ \LARGE\Rightarrow I=I+1\Rightarrow 1=0$

Which is (of course)wrong.

What we should see here is the importance of the arbitrary constant of integration $\Large C$.

So if the constant of integration on the left hand side is $\Large C_1$ and on the right hand side the constant is $\Large C_2$, we may see that

$\LARGE C_1-C_2=1$

Which resolves the fallacy.

We might well see that the functions on both the right and left hand side have the same derivative i.e. $\frac{1}{x\ln(x)}$ ,proving that we are right.

We might have many interesting apparent "fallacies" as mentioned in this post .

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