Question

evaluate the following integral.
int (2xe^{x}^{2})/(1+e^{x}^{2})

Answer 1

\[\int\limits_{}^{} 2xe ^{x ^{2}}/1+e ^{x ^{2}}\]

Answer 2

the top is the derivative of the bottom right?

Answer 3

this is what I have

Answer 4

\[\int\frac{dx}{x}=ln|x|+C\]

Answer 5

\[\int\limits_{}^{} 2xe ^{x ^{2}}/1+e ^{x ^{2}}dx\]
\[\int\limits_{}^{} 2e ^{x ^{2}}x/1+e ^{x ^{2}}\]
\[2\int\limits_{}^{} e ^{x ^{2}}x/1+e ^{x ^{2}}\]
\[u=e ^{x ^{2}}\]
\[du=2e ^{x ^{2}}\]
\[\int\limits_{}^{} 1/u\]
\[\ln(u)+c\]
\[\ln(e ^{x ^{2}}+1)+c\]

Answer 6

the problem i am having trouble with is the step right before i integrate \[2\int\limits_{}^{}x^{x ^{2}}x/1+e ^{x ^{2}}\] does the top portion of the equation equal 1?????

Answer 7

in the case above; u = e^(x^2); du = 2x e^(x^2)

Answer 8

that first x is supposed to be an e

Answer 9

why are you pulling out the 2?

Answer 10

because the 2 is a constant

Answer 11

just becasue it is a constant doent not mean you pull it out; it is very useful to us so leave it in

Answer 12

so i was under the impression to always pull out the constant. how do i know when and when not to pull out the constant?

Answer 13

du \(2x\ e^{x^2}\)
--- = -------
u \(1+e^{x^2}\)

you only pull out a constant if its useless to the over all integration

Answer 14

the 2 helps to establish the fact that the top is the derivative of the bottom

Answer 15

we know how to integrate du/u .... so ues it :)

Answer 16

i see. that helps clear things up. thanks

Answer 17

youre welcome