Why don't we add a constant of integration when integrating the second function while applying integration by parts? I'm searching for a perfectly logical answer

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestWhat is "the second function"? On \(\int u \, dv = uv - \int v \, du\), is the second function \(v\) or \(v \, du\)? – Ivan Koswara · 1 year, 3 months ago

Log in to reply

– Sanchit Aggarwal · 1 year, 3 months ago

The second function is 'dv', the function whose integral is 'v'Log in to reply

It will not affect the final answer. You can try it out yourself. The extra part, if you think there will be any, will get cancelled out at the end. – Sanchit Aggarwal · 1 year, 3 months ago

Log in to reply

P.S. You have to integrate g(x) twice while applying by parts – Mayank Singh · 1 year, 3 months ago

Log in to reply

P.s. refer to the derivation of the formula once – Sanchit Aggarwal · 1 year, 3 months ago

Log in to reply

Thanks for asking this question, I had the same one in my mind! – Swapnil Das · 1 year, 3 months ago

Log in to reply

– Mayank Singh · 1 year, 3 months ago

Well I've reasoned something out... I'm waiting for some other repliesLog in to reply

Log in to reply

– Mayank Singh · 1 year, 3 months ago

Just last yearLog in to reply