Question

Question about integral of arctan(2x)dx

Answer 1

So I got it down to the point where it's \[\int\limits \frac{2x}{1+4x^2}\]

Answer 2

or \[\int\limits \frac{2x}{1+(2x)^2}\]

Answer 3

now at first I thought I could cancel out the 2x, can I not do that?

Answer 4

because I got 1/2 ln|1+2x| and theirs was 1/4 ln|1+4x^2|

Answer 5

let u = 2x
du = 2dx
\[\implies \frac 12 \int arctan (u) du\]
let w = arctan u
dw = 1/u^2 + 1

dv = du
v = u

\[wv - \int vdw\]
\[\implies u \tan^{-1} u - \int \frac{u}{u^2 + 1}du\]
now do trig sub

does that help?

Answer 6

you did it a much different way lol...

Answer 7

weird way boo iggy....

Answer 8

works tho....

Answer 9

Anyways dood...... Can I cancel the 2x from the (2x)^2 or not?

Answer 10

\[\tan \theta = u\]
\[\sec^2 \theta d\theta = du\]
\[\sec\theta = \sqrt{u^2 + 1}\]
\[\sec^2 \theta = u^2 + 1\]
\[\implies \int \frac{\tan \theta \sec^2 \theta d\theta}{\sec^2 \theta}\]
\[\implies \int \tan \theta d\theta\]
\[\implies \ln | \sec \theta|\]
\[\implies \ln | \sqrt{u^2 + 1}|\]
okay this is sooo long

Answer 11

no you cant

Answer 12

ok that's all I need to do.. and WTF ARE YOU DOING DOOD LOL.

Answer 13

you'll get there....someday...lol

Answer 14

all I needed to know*

Answer 15

and i typed all that =_=

Answer 16

leave me alone :'(.

Answer 17

Dude I said what I needed from the start :).

Answer 18

really? must've missed it

Answer 19

so why can't I cancel it? 2x/(1+(2x)^2)?

Answer 20

why not to try substitution
let
t=1+4x^2
dt=8xdx
1/8dt=xdx
??

Answer 21

it's like \[\frac{a}{1+a^2}\]
there's an addition...you can only cancel when it's multiplication like \[\frac{a}{ab}\]

Answer 22

Yup Sami that's what I was doing.

Answer 23

anyway... how come you just got \[\int \frac{2x}{1 + (2x)^2} dx\]

Answer 24

But at first I tried to cancel, then I realized it's a du/u and then did it that way. When I went to check my answer they didn't cancel, and I didn't think about doing du/u before that.

Answer 25

you used integration by parts right?

Answer 26

mhm

Answer 27

you used u = arctan (2x) and dv = dx?

Answer 28

yessir.

Answer 29

so where is the arctan in your solution?

Answer 30

the entire solution is xarctan(2x) - 1/4ln|1+4x^2|

Answer 31

the only part I was concerned with was why I couldn't cancel 2x from (2x)^2.. :'(?

Answer 32

it was algebra

Answer 33

algebra hindered you

Answer 34

anyway...just so you know what we did are just same...that's why i told you you'll get there..in the end lol

Answer 35

IT aslways does bro I didn't have any teachers for all of middle school.

Answer 36

I know.... I'll just have to remember I cannot cancel like that... Idk why tho
explain1

Answer 37

can i use numbers to show you?