Question

interal (dx/sqrt(x^(2)+16)

Answer 1

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{x^2+16}~dx}\) ?

Answer 2

correct interpretation ?

Answer 3

\[\int\limits \frac{ dx }{ \sqrt{x^2+16} }\] like this

Answer 4

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\sqrt{x^2+16}}~dx}\)

Answer 5

Yeah

Answer 6

First, factor the 16 out of the radical.

Answer 7

what ?

Answer 8

nope

Answer 9

this is one of those trig integrals

Answer 10

yeah is the form \[\sqrt{x^2+a^2}\]

Answer 11

HI!!

Answer 12

so what did you do so far?

Answer 13

try \(x=4\tan(\theta)\)

Answer 14

the radical will go bye bye lickety split

Answer 15

yeah is x=aTan(angle), so as @misty1212 says a=4 and x=4Tan(theta)

Answer 16

well i guess this is solved :)

Answer 17

the person that asked this question left :P and we're still trying to help *sighs*

Answer 18

\[\int\limits_{}^{}\frac{ 1 }{ 4\sqrt{(\frac{ x }{ 4 }})^{2}+1 }\]

Answer 19

@NoelGreco a direct sub method is the way to go

Answer 20

for a simpler method, memorize
\[\frac{d}{dx}[\sinh^{-1}(x)]=\frac{1}{\sqrt{1+x^2}}\]

Answer 21

eh i'm against memorizing hehe
i was going to say that arcsin' = that stuff but better learn that sub

Answer 22

Now you have it in the form x= tan theta, d/d theta =sec^2 theta.

Answer 23

you can prove that what mitsy posted as well, if you like very simply.

Answer 24

I do like memorizing:D

Answer 25

why did jacket leave?

Answer 26

this user is probably south from the equator jk

Answer 27

hehe
good analysis :)

Answer 28

@SolomonZelman how to use show that one
i haven't seen that much of arcsinh
it just kind of remind me of arctan just the square root difference

Answer 29

how do you*

Answer 30

you want arctan(s) or what?

Answer 31

arctan(x) * ?

Answer 32

\(\large\color{slate}{ \displaystyle \frac{d}{dx}\tan^{-1}x=1/( x^2+1) }\)
~~~~~~~~~~~~~~~~~~~~~~~~~

\(\large\color{slate}{ \displaystyle y=\tan^{-1}x }\)

\(\large\color{slate}{ \displaystyle \tan y=x }\)

derivative,

\(\large\color{slate}{ \displaystyle y'~\sec^2y=1 }\)

\(\large\color{slate}{ \displaystyle y'~(\tan^2y+1)=1 }\)

\(\large\color{slate}{ \displaystyle y'~( x^2+1)=1 }\)

\(\large\color{slate}{ \displaystyle y'=1/( x^2+1) }\)

Answer 33

this simple thing you want?

Answer 34

who is here? was it the world's biggest lag? jk

Answer 35

igtg in a couple minutes it was nice to see the problem.

Answer 36

no i meant arcsinh i know arctan of course :)

Answer 37

i thought you were talking about arcsinh'

Answer 38

@SolomonZelman