# 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
3 years, 1 month 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.

- 3 years, 1 month ago

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

- 3 years, 1 month ago

- 3 years, 1 month ago

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

- 3 years, 1 month ago

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

- 3 years, 1 month ago

I can't believe you just stated that...

- 3 years, 1 month ago

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

- 3 years, 1 month ago

its $\frac{b}{b}-\frac{a}{a}$ and still $1|_a^b = 1_a-1_b=1-1=0$.

- 3 years, 1 month ago

Oops! I can't believe I made such a mistake... Anyways thx..

- 3 years, 1 month ago

- 3 years, 1 month ago

Can you post this as a problem so others can learn from this misconception?

Staff - 3 years, 1 month ago

Ok, surely

- 3 years, 1 month ago

Exactly... Then he should get correctly.

- 3 years, 1 month ago

How is LHS not equal to 0

- 3 years, 1 month ago

Integration constants need not be equal.

- 3 years, 1 month ago

K.. I agree on that

- 3 years, 1 month ago

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

- 3 years, 1 month ago

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

- 3 years, 1 month ago

Oh, right, thank you

- 3 years, 1 month ago

Welcome.. :)

- 3 years, 1 month ago

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.

- 2 years, 2 months ago