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Is 0=1? Then find the mistake

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

Note by Sparsh Sarode
6 months, 2 weeks ago

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you actually got \(\int \frac{1}{x} -\int \frac{1}{x}=1 \). There is absolutely nothing wrong with that. Those are indefinite integrals. Kushal Patankar · 6 months, 2 weeks ago

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@Kushal Patankar How is LHS not equal to 0 Sparsh Sarode · 6 months, 2 weeks ago

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@Sparsh Sarode Integration constants need not be equal. Kushal Patankar · 6 months, 2 weeks ago

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@Kushal Patankar K.. I agree on that Sparsh Sarode · 6 months, 2 weeks ago

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@Kushal Patankar I should have added limits. Sparsh Sarode · 6 months, 2 weeks ago

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@Sparsh Sarode It is \(\frac{x}{x} |_{a}^{b}\). That yeilds zero. 1 there has limits. Kushal Patankar · 6 months, 2 weeks ago

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@Kushal Patankar Exactly... Then he should get correctly. Rishabh Cool · 6 months, 2 weeks ago

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@Kushal Patankar I didn't quite get u, u meant \( \dfrac{x}{x} =\dfrac{b-a}{b-a} \neq 1\) Sparsh Sarode · 6 months, 2 weeks ago

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@Sparsh Sarode I can't believe you just stated that... Kushal Patankar · 6 months, 2 weeks ago

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@Kushal Patankar Correct me if I am wrong, isn't \( \dfrac{2-1}{2-1}=\dfrac{1}{1}=1\) ? Sparsh Sarode · 6 months, 2 weeks ago

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@Sparsh Sarode its \(\frac{b}{b}-\frac{a}{a}\) and still \(1|_a^b = 1_a-1_b=1-1=0\). Kushal Patankar · 6 months, 2 weeks ago

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@Kushal Patankar Oops! I can't believe I made such a mistake... Anyways thx.. Sparsh Sarode · 6 months, 2 weeks ago

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@Sparsh Sarode Can you post this as a problem so others can learn from this misconception? Calvin Lin Staff · 6 months, 2 weeks ago

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@Calvin Lin Ok, surely Sparsh Sarode · 6 months, 2 weeks ago

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@Sparsh Sarode Sure... Glad to help. Kushal Patankar · 6 months, 2 weeks ago

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@Kushal Patankar What if I add limits from 1 to 2..you still get 1=0 Sparsh Sarode · 6 months, 2 weeks ago

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And what happens without limits? \[\displaystyle \int \frac{dx}{x} =\frac{x}{x}+\displaystyle \int \frac{dx}{x}\] Hjalmar Orellana Soto · 6 months, 2 weeks ago

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@Hjalmar Orellana Soto It's still correct.. Integration constants need not be same Sparsh Sarode · 6 months, 2 weeks ago

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@Sparsh Sarode Oh, right, thank you Hjalmar Orellana Soto · 6 months, 2 weeks ago

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@Hjalmar Orellana Soto Welcome.. :) Sparsh Sarode · 6 months, 2 weeks ago

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