Consider the integral, \(\displaystyle I =\int \dfrac{1}{x}dx \)

We know that \( I= \log x \)

But, \(\displaystyle I =\int \dfrac{1}{x}(1)dx =(x)\dfrac{1}{x}-\int \bigg(-\dfrac{1}{x^2} \bigg) xdx\)

\(1+I=I \Rightarrow 1=0\)

What is the mistake?

I do not know the mistake

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## Comments

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TopNewestyou actually got \(\int \frac{1}{x} -\int \frac{1}{x}=1 \). There is absolutely nothing wrong with that. Those are indefinite integrals.

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How is LHS not equal to 0

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Integration constants need not be equal.

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I should have added limits.

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It is \(\frac{x}{x} |_{a}^{b}\). That yeilds zero. 1 there has limits.

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What if I add limits from 1 to 2..you still get 1=0

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You cannot cancel the integrals on both sides because each function has infinitely many anti-derivatives off by a constant. That's why, for good reason, you should always include "+C" in your answers.

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And what happens without limits? \[\displaystyle \int \frac{dx}{x} =\frac{x}{x}+\displaystyle \int \frac{dx}{x}\]

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It's still correct.. Integration constants need not be same

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Oh, right, thank you

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