Question

d/dx int (u^2-1)/(u^2+1) from 2x to 3x

Answer 1

\[d/dx \int\limits (u^2-1)/(u^2+1) from 2x \to 3x\]

Answer 2

... i forgot something lol

Answer 3

\[\frac{d}{dx}\left(\int_{2x}^{3x}(\frac{u^2-1}{u^2+1})\right)\]

Answer 4

Yep

Answer 5

well, there is a way to use the fundamental thrm to this, but I never pay attention to it, makes life way to easy :)

Answer 6

I end up going through the calculations and getting \[(3* (3x-1)/(3x^2+1)) + (2* (2x^2-1)/(2x^2+1)) \]

Answer 7

3x^2*

Answer 8

it ints up to:
\[F(3x)-F(2x)\] and then deriving back down to this little equation

Answer 9

the F(x) is really inconsequantial to the process; but I cant recall how

Answer 10

\[\int_{2x}^{3x}f'(u)du \implies f(3x)-f(2x)\]
\[\frac{d}{dx}(f(3x)-f(2x)) \implies f'(3x)*3 - f'(2x)*2\]
\[f'(u)=\frac{u^2-1}{u^2+1}\]
\[\frac{(3x^2)-1}{(3x^2)+1}-\frac{(2x^2)-1}{(2x^2)+1}\]
right?

Answer 11

ack... all that work and I left out the *3 and *2 lol

Answer 12

The answer turns out be [according to the solution manual [(-2 * 4x^2-1/4x^2+1)+(3* 9x^2-1/9x^2+1)\]

Answer 13

\[\frac{3((3x)^2-1)}{(3x)^2+1}-\frac{2((2x)^2-1)}{(2x)^2+1}\]

Answer 14

yeah, when I get all the parts right; I see it happening :)

Answer 15

My question is where does the second 3 and 2 come from that you multiply.

Answer 16

the chain rule really....

Answer 17

but, the 2 and 3 are already in the equation and aren't being multiplied..

Answer 18

it doesnt really matter what the function integrates up to; because you are simply bringing it back down to the derivative again right?

Answer 19

Yea, it's just putting it into terms of x .

Answer 20

Oh snap, never mind I see it.

Answer 21

\[F(x)=\int_{2x}^{3x}\left(\frac{u^2-1}{u^2+1}\right)du\]
and then you work in the \(F(3x)-F(2x)\) and derive it back again

Answer 22

Herp derp, thanks a bunch.

Answer 23

:) youre welcome