Question

Someone please help

Answer 1

@zepdrix

Answer 2

Find integral of arctan(1/x)Dx from 1 to square root 3

Answer 3

Umm so how do we start?
Probably parts, ya?

Answer 4

Yeah :(

Answer 5

\(\large\rm u=arctan\left(\frac{1}{x}\right)\qquad\qquad\qquad dv=dx\)

Answer 6

Omg :(

Answer 7

I always put like cos or sin or whatever and than the inside as the other part :(

Answer 8

Oh, like you accidentally put dv=1/x dx or something like that?

Answer 9

Yes

Answer 10

Ya, that's no bueno.
Find your \(\large\rm du\) and \(\large\rm v\).

Answer 11

Okay

Answer 12

So v=x and du=1/(1+1/x^2))

Answer 13

No, you forgot to chain rule on your du.

Answer 14

\[\large\rm \left(\arctan\frac{1}{x}\right)'\quad=\quad\frac{1}{1+\left(\frac{1}{x}\right)^2}\left(\frac{1}{x}\right)'\]

Answer 15

Ughhhh

Answer 16

Calc is a struggle

Answer 17

The struggle is real.

Answer 18

It is

Answer 19

Derivative of 1/x is -x^-2

Answer 20

-1/x^2
ok good.\[\large\rm \left(\arctan\frac{1}{x}\right)'\quad=\quad\frac{-\frac{1}{x^2}}{1+\left(\frac{1}{x}\right)^2}\]We should simplify this before proceeding,
any ideas?

Answer 21

Hmmmm

Answer 22

well we can square than cancel?

Answer 23

We don't like fractions on top of fractions.
So we would like to multiply the numerator and denominator by the `Least Common Multiple` of these denominators.

It turns out that both denominators are x^2, ya?

Answer 24

Yes!

Answer 25

\[\large\rm \left(\arctan\frac{1}{x}\right)'\quad=\quad\frac{-\frac{1}{x^2}}{1+\frac{1}{x^2}}\color{royalblue}{\frac{x^2}{x^2}}\]So give an x^2 to the numerator and denominator.

Answer 26

Okay

Answer 27

Understand how that will get us to this point?\[\large\rm \left(\arctan\frac{1}{x}\right)'\quad=\quad\frac{-1}{x^2+1}\]

Answer 28

Yes

Answer 29

\[\large\rm \int\limits_1^{\sqrt3}\arctan\left(\frac{1}{x}\right)dx\]And we've chosen our parts as such,\[\large\rm u=\arctan\left(\frac{1}{x}\right)\qquad\qquad\qquad dv=dx\]\[\large\rm du=\frac{-1}{1+x^2}dx\qquad\qquad\quad\qquad v=x\]

Answer 30

Yes :)

Answer 31

Ok plug in the stuff :o
do the things

Answer 32

Lol I wish I had equation

Answer 33

Arctan(1/x)(x) Integral (x)(-1/x^2+1)

Answer 34

Ok let's combine the negatives,\[\large\rm x \arctan\left(\frac{1}{x}\right)+\int\limits\frac{x}{1+x^2}dx\]

Answer 35

So uh, how bout that new integral?
Any ideas?

Answer 36

Well move the (x^2+1) to the top to be a negative power?

Answer 37

And then what? Parts again?
That doesn't seem great.

Answer 38

Hmmm

Answer 39

Wait

Answer 40

1/1+x^2 is arctan?

Answer 41

Yes, but that would occur by doing parts...
and that would effectively `undo` the parts that we first did.

So that idea is not going to work.

Answer 42

I give up

Answer 43

u-substitution.

Answer 44

Ohhhhh yeah

Answer 45

Okay

Answer 46

Will you get ln in this?

Answer 47

1/2 integral ln(x^2+1)?

Answer 48

Oops no integral

Answer 49

Ya that looks great.\[\large\rm x \arctan\left(\frac{1}{x}\right)+\frac{1}{2}\ln|1+x^2|\]Now we need to evaluate this at our limits.

Answer 50

Fun :/

Answer 51

I don't wanna type it :/

Answer 52

Square root 3arctan(1/square root 3)+1/2ln|1+(square root 3)^2

Answer 53

lol

Answer 54

I don't wanna type it either -_-
I'll draw it I guess

Answer 55

Okay :)