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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
2 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. · 2 months, 2 weeks ago

How is LHS not equal to 0 · 2 months, 2 weeks ago

Integration constants need not be equal. · 2 months, 2 weeks ago

K.. I agree on that · 2 months, 2 weeks ago

I should have added limits. · 2 months, 2 weeks ago

It is $$\frac{x}{x} |_{a}^{b}$$. That yeilds zero. 1 there has limits. · 2 months, 2 weeks ago

Exactly... Then he should get correctly. · 2 months, 2 weeks ago

I didn't quite get u, u meant $$\dfrac{x}{x} =\dfrac{b-a}{b-a} \neq 1$$ · 2 months, 2 weeks ago

I can't believe you just stated that... · 2 months, 2 weeks ago

Correct me if I am wrong, isn't $$\dfrac{2-1}{2-1}=\dfrac{1}{1}=1$$ ? · 2 months, 2 weeks ago

its $$\frac{b}{b}-\frac{a}{a}$$ and still $$1|_a^b = 1_a-1_b=1-1=0$$. · 2 months, 2 weeks ago

Oops! I can't believe I made such a mistake... Anyways thx.. · 2 months, 2 weeks ago

Can you post this as a problem so others can learn from this misconception? Staff · 2 months, 1 week ago

Ok, surely · 2 months, 1 week ago

Sure... Glad to help. · 2 months, 2 weeks ago

What if I add limits from 1 to 2..you still get 1=0 · 2 months, 2 weeks ago

And what happens without limits? $\displaystyle \int \frac{dx}{x} =\frac{x}{x}+\displaystyle \int \frac{dx}{x}$ · 2 months, 2 weeks ago

It's still correct.. Integration constants need not be same · 2 months, 2 weeks ago

Oh, right, thank you · 2 months, 2 weeks ago