An apparent fallacy

If we want to integrate

\(\LARGE \int \frac{dx}{xlnx}\)

We may use the integration by parts technique instead of substitution to get I=∫dxxln⁡x=1ln⁡x∫dxx−∫(−1(ln⁡x)21x∫dxx)dx=1ln⁡xln⁡x+∫dxxln⁡x=1+I⇒I=I+1⇒1=0 \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 C\Large C.

So if the constant of integration on the left hand side is C1\Large C_1 and on the right hand side the constant is C2\Large C_2, we may see that

C1−C2=1\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. 1xln⁡(x)\frac{1}{x\ln(x)} ,proving that we are right.

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

#Calculus #IntegrationByParts #IntegrationTechniques

Note by Krishna Jha
6 years, 7 months ago

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