Question

Integrate (x)/(1+x^2) DX ....show work please

Answer 1

u - sub for this one.

put
\[u=1+x^2, du=2xdx,\frac{1}{2}du = dx\] and do the rest mostly in your head

Answer 2

http://www.wolframalpha.com/input/?i=1%2F2+log%281%2Bx^2%29&lk=1&a=ClashPrefs_*Math-

There is full calculation more than i can explain ... =)

Answer 3

\[\int\frac{du}{u}=\ln(u)\] replace by u by
\[x^2+1\] and dont forget the one half out front and +C at the end

Answer 4

Srry this is the one.. http://www.wolframalpha.com/input/?i=%28x%29%2F%281%2Bx^2%29

Answer 5

Thanks satellite, I'm still not completely comfortable with the application of U.

Answer 6

did you get this one?

Answer 7

Yes I understand it, the only part of integration that becomes confusing for me is selecting which part of the equation to turn into U.

Answer 8

ah i see. it will make more sense if you take the derivative of your answer and see why it works. "u-sub" just a gimmick for doing the chain rule backwards. so usually you use it when you see a composite function times its derivative.

Answer 9

hmm, trying to do equation....still a bit lost here.

Answer 10

if you take the derivative of
\[\ln(x^2+1)\] you see that you get
\[\frac{1}{x^2+1}\times 2x=\frac{2x}{x^2+1}\] by the chain rule. this is almost exactly what you want, you are just off by a factor of 2

Answer 11

that is what the "u-sub' allows you to do. see what you need to adjust by to get the right thing

Answer 12

how did you get to ln(x^2 + 1)

Answer 13

ok lets to slow

Answer 14

your job is
\[\int\frac{x}{x^2+1}dx\] meaning you are looking for some function whose derivative is
\[\frac{x}{x^2+1}\]

Answer 15

in the denominator you have
\[x^2+1\] and in the numerator you have
\[x\] which is almost the derivative of the denominator. you are only off by a factor of 2

Answer 16

this is a set up for a u - substitution, which is device for running the chain rule backwards.
so you can say:
put
\[u=x^2+1\] and then
\[du=2xdx\]
so
\[\frac{1}{2}du=dx\] and now rewrite your integral entirely in terms of u as
\[\frac{1}{2}\int\frac{1}{u}du\]

Answer 17

thats where im confused

Answer 18

what does this buy you? well now you know exactly an anti-derivative of
\[\frac{1}{u}\] namely
\[\ln(u)\] because
\[\frac{d}{du}\ln(u)=\frac{1}{u}\]

Answer 19

why isnt it x/u instead of 1/u

Answer 20

sorry i meant you have written
\[2xdx=du\implies xdx=\frac{1}{2}du\]

Answer 21

ahhhhh, X's cancel out?

Answer 22

so now you have everything in terms of u. no they don't "cancel" they just disappear. it is a trick, a trick to write your integral in a form you can answer

Answer 23

again let me stress this is just a gimmick to get what you want. you want a function whose derivative is
\[\frac{x}{x^2+1}\] so how are you going to find it? once we recognize that he denominator is almost the derivative of the numerator we know that it is going to be
\[\ln(x^2+1)\] but we need to know how to adjust.

Answer 24

multiply by the 1/2 left over and its your answer right?

Answer 25

right!

Answer 26

here is another maybe simpler example. suppose you want
\[\int x\cos(x^2)dx\]

Answer 27

u= x^2?

Answer 28

so you know that the derivative of sine is cosine and you are almost home free. you know your answer will be
\[\sin(x^2)\] basically. but the derivative of
\[\sin(x^2)\] is
\[2x\cos(x)\] which is not exactly what you have. so you have to divide by two.
you are right. you would say
\[u=x^2\]
\[du=2xcx\]
\[\frac{1}{2}du=xdx\] and now you have exactly what you want

Answer 29

ok thanks, I actually havent seen sin or cos yet in calc, but prob will soon

Answer 30

oh well good luck! and if you get confused check your answer by differentiation

Answer 31

thankfully I have the ti 89 to help me out as well haha, thanks