# 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
1 year, 11 months 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.

- 1 year, 11 months ago

How is LHS not equal to 0

- 1 year, 11 months ago

Integration constants need not be equal.

- 1 year, 11 months ago

K.. I agree on that

- 1 year, 11 months ago

I should have added limits.

- 1 year, 11 months ago

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

- 1 year, 11 months ago

Exactly... Then he should get correctly.

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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

- 1 year, 11 months ago

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

Staff - 1 year, 11 months ago

Ok, surely

- 1 year, 11 months ago

Sure... Glad to help.

- 1 year, 11 months ago

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

- 1 year, 11 months 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.

- 1 year ago

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

- 1 year, 11 months ago

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

- 1 year, 11 months ago

Oh, right, thank you

- 1 year, 11 months ago

Welcome.. :)

- 1 year, 11 months ago