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

## 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. – Kushal Patankar · 8 months, 1 week ago

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– Sparsh Sarode · 8 months, 1 week ago

How is LHS not equal to 0Log in to reply

– Kushal Patankar · 8 months, 1 week ago

Integration constants need not be equal.Log in to reply

– Sparsh Sarode · 8 months, 1 week ago

K.. I agree on thatLog in to reply

– Sparsh Sarode · 8 months, 1 week ago

I should have added limits.Log in to reply

– Kushal Patankar · 8 months, 1 week ago

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

– Rishabh Cool · 8 months, 1 week ago

Exactly... Then he should get correctly.Log in to reply

– Sparsh Sarode · 8 months, 1 week ago

I didn't quite get u, u meant \( \dfrac{x}{x} =\dfrac{b-a}{b-a} \neq 1\)Log in to reply

– Kushal Patankar · 8 months, 1 week ago

I can't believe you just stated that...Log in to reply

– Sparsh Sarode · 8 months, 1 week ago

Correct me if I am wrong, isn't \( \dfrac{2-1}{2-1}=\dfrac{1}{1}=1\) ?Log in to reply

– Kushal Patankar · 8 months, 1 week ago

its \(\frac{b}{b}-\frac{a}{a}\) and still \(1|_a^b = 1_a-1_b=1-1=0\).Log in to reply

– Sparsh Sarode · 8 months, 1 week ago

Oops! I can't believe I made such a mistake... Anyways thx..Log in to reply

– Calvin Lin Staff · 8 months, 1 week ago

Can you post this as a problem so others can learn from this misconception?Log in to reply

– Sparsh Sarode · 8 months, 1 week ago

Ok, surelyLog in to reply

– Kushal Patankar · 8 months, 1 week ago

Sure... Glad to help.Log in to reply

– Sparsh Sarode · 8 months, 1 week ago

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

And what happens without limits? \[\displaystyle \int \frac{dx}{x} =\frac{x}{x}+\displaystyle \int \frac{dx}{x}\] – Hjalmar Orellana Soto · 8 months, 1 week ago

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– Sparsh Sarode · 8 months, 1 week ago

It's still correct.. Integration constants need not be sameLog in to reply

– Hjalmar Orellana Soto · 8 months, 1 week ago

Oh, right, thank youLog in to reply

– Sparsh Sarode · 8 months, 1 week ago

Welcome.. :)Log in to reply