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\[\int\limits2b(1+t)dt\]
How does my prof get b(1+t)^2 + C but I got 2bt +bt^2 +C

\[2b*t+2b*\frac{t^{2}}{2} +c\]
\[2bt+bt^2+c\]
it is the same..

your professor is right .
because
\[\Large \int\limits_{}^{} (ax+b)=\frac{(ax+b)^2}{a}+C\]

its 2a !

k is a constant. Just replace that k with a big C.

b(1+t)^2 = b(1+2t+t^2)=b+2bt+bt^2
but this is not the same as 2bt+bt^2 o_O

but im still right!

am I missing something?

yes you are !

@kirbykirby You meant to differentiate it.

You expand then differentiate.

Wait sorry you differentiate by not expanding.

differentiate?? this is integration!

No, he wants to get the answer back to the original expression.

Ok. I'm trying to understand though how the 2 methods though give a different answer though...

differentiate b+2bt+bt^2 with respect to t and you get:
2b+2bt=2b(1+t)

Which is the original expression you had in the question...

oh is it because the b + C gets absorbed into another constant "C" using the other method o_O?

isn't it a constant here with respect to t?

Yes.

If it was respect to b, then it's a different story.

integrate*

\[2bt+bt^2 \neq=2b(t+1)\]

oops forget the extra "2" in the last expression

I am not learning how to integrate :( I know how to integrate

I just don't see how these 2 methods are giving different answers

first put*

x^2*

Then use one method. Simple as that.

But why does one not give the same answer as the other :( THIS is what I wanna understand

If you really want to know, then go to your professor and ask him why.

Your professor's there to help you understand, not bore you to death.

*ask him/her