Question

find the integral... 17arctan(x) dx

Answer 1

pull out the 17 and then use your integration table

Answer 2

\[17[\frac{1}{1 + x^2} ]+ c\]

Answer 3

\[\frac{17}{1 + x^2} + c\]

Answer 4

I think this is correct? :P

Answer 5

hmm guess not... Then would it be 17 on the bottom also?

Answer 6

all you have to do is prove that arctan is the derivative of what you have

Answer 7

or you can do byparts if you want since that is how it's derived i believe

Answer 8

I didn't think we needed by parts for this....

Answer 9

it's how it's derived i believe let me see

Answer 10

It seems like it though, wolfram has a weird answer like that so I guess so...

Answer 11

so u = arctan(x)

\[du = \frac{1}{1 + x^2}\]

dv = dx

v = x

Answer 12

I can pul the 17 out right? or would it be v = 17x?

Answer 13

dv=dx
v=x

u=arctan(x)
\[du=\frac{1}{1+x^2}\]

\[\int udv=uv-\int vdu\]

Answer 14

you can pul it out it's a constant

Answer 15

just make sure it multiplies everything

Answer 16

\[xarctan(x) - \int\limits \frac{x}{1+x^2} dx\]

Answer 17

correct now can yu do the second intergral

Answer 18

\[17[xarctan(x) - \int\limits \frac{x}{1+x^2} dx]\]

Answer 19

so in this case u = 1+x^2
du = x

\[\int\limits \frac{du}{u} \]

Answer 20

\[17xarctan(x) - 17\log(u) + c\]

Answer 21

du=2xdx

Answer 22

\[17xarctan(x) - 17\log(1+x^2) + c\]

Answer 23

woopsies :P

Answer 24

yeah

Answer 25

\[17xarctan(x) - \frac{17}{2}\log(1+x^2) + c\]

Answer 26

sorry huge headache and doing this too fast :P.

Answer 27

so apparently that's not right hmmph.

Answer 28

that's not the right answer?

Answer 29

grr it wanted LN WTF!@@!~@~

Answer 30

that's a bummer ask your teacher about it

Answer 31

It's all good, still got it right, but still booo!

Answer 32

I keep forgetting about LN...

Answer 33

Can i ask a question

why isn't 17/(1+x^2)+C the answer ? ?

Answer 34

That's what I wanted to know lol :P.

Answer 35

I guess because that's only the derivative...

Answer 36

beause the derivative of arctan(x) is u'/1+u^2

Answer 37

you'd have to have something such as du/a^2+u^2 to do this

Answer 38

ugly...

Answer 39

@Outkast3r09 OMG!! Yes

Thanksss :D

Answer 40

yupp...

Answer 41

the answer would be an arctan...

Answer 42

Uh-oh!!! my brain stopped working lol

Answer 43

Same!

Answer 44

if it was like that... the parts method would have brought us back to it eventually depending on whether we needed to simplify or not