Question

Calculas problem...

Answer 1

\[\frac{ d }{ dx } \int\limits_{2x}^{3x} (\frac{ u^2-1 }{ u^2 + 1 }) du\]

Answer 2

So we don't really need to find the antiderivative
So the antiderivative of the f(u)=(u^2-1)/(u^2+1)
is F(u)
\[\frac{d}{dx}\int\limits_{2x}^{3x}\frac{u^2-1}{u^2+1} du \\ =\frac{d}{dx}(F(u)|_{2x}^{3x})\]
so now plug in your limits

Answer 3

\[=\frac{d}{dx}(F(3x)-F(2x))\]
and differentiate :)

Answer 4

you must use chain rule

Answer 5

well forget it ;) now i got it :)

Answer 6

thank you very much :)

Answer 7

So I guess that means you are good with differentiating the F(3x) and the F(2x) thing?

Answer 8

yes but i have studied substitution rule and i got a little confused ! i thought i should use it

Answer 9

so you wanted to try to also integrate it?

Answer 10

like the part before the differentiating part

Answer 11

correct ! and now you gave me an idea which is better :) it would be very confusing in that way.

Answer 12

u can differenntiatite first and then integrate too
q

Answer 13

q:

Answer 14

:Pq:

Answer 15

nvm not in this case

Answer 16

\[\int\limits\limits_{}^{}\frac{u^2-1}{u^2+1} du=\int\limits\limits_{}^{}\frac{u^2+1-2}{u^2+1}du=\int\limits\limits_{}^{}(1-\frac{2}{u^2+1}) du=u-2 \arctan(u)+C \]
\[\int\limits_{2x}^{3x}f(u)du=3x-2\arctan(3x)-2x+2 \arctan(2x) \\ \frac{d}{dx} \int\limits_{2x}^{3x}f(u) du=3-2 (3) \frac{1}{1+(3x)^2}-2+2(2)\frac{1}{1+(2x)^2}\]
and if we have done it the other way we would have:
\[
\frac{d}{dx}(F(3x)-F(2x))=3f(3x)-2f(2x) \\ =3\frac{(3x)^2-1}{(3x)^2+1}-2 \frac{(2x)^2-1}{(2x)^2+1}
\]

Answer 17

yes it was 3 ... thank you very much i guess the second way is better ;)

Answer 18

:) too easy

Answer 19

its too easy man

Answer 20

\[=3 \frac{(3x)^2+1-2}{(3x)^2+1}-2\frac{(2x)^2+1-2}{(2x)^2+1} \\ =3(1-\frac{2}{(3x)^2+1})-2(1-\frac{2}{(2x)^2+1}) \\ =3-\frac{3(2)}{(3x)^2+1}-2+\frac{2(2)}{(2x)^2+1}\]
and as we see we get the same thing

Answer 21

I don't know
the first way might be easier :p

Answer 22

like the second way we actually had to integrate

Answer 23

no need for any work u just plug in

Answer 24

3x and 2x

Answer 25

the differentiation and the integration cancel out
but u have to evaluate at right bounds

Answer 26

d/dx of integral bounded a(x) to b(x) of f(x) dx=f(a(x))-f(b(x))

Answer 27

@myininaya Quick question: How did you go from...\[\int\frac{u^2+1-2}{u^2+1}du \to \int\left(1-\frac{2}{u^2+1}\right)du\]

Just that little portion is confusing me :)

Answer 28

\[\frac{u^2+1}{u^2+1}-\frac{2}{u^2+1}\]