Question

\[\int \frac{dx}{\sqrt{1+x^2}}\]

Answer 1

I know that \(\dfrac{1}{1+x^2} = \dfrac{d}{dx}(\tan^{-1}(x))\)

Answer 2

:D

Answer 3

for fun yayyy

Answer 4

why so , u can post several questions :P

Answer 5

:D...

Answer 6

ok i'll work on complex instead of regular way

Answer 7

you can do trig substitution
x = tan(theta)

Answer 8

nope.

Answer 9

Makes it more difficult than it needs to be.

Answer 10

lol i know its only easy sub but since its for fun i wanted to try something else xD

Answer 11

Hyperbolic trig subs. \[d(\sinh x) = \frac{1}{\sqrt{1+x^2}}\]

Answer 12

Make it hard, use polar. Haha

Answer 13

Go ahead @Marki :P

Answer 14

haha ok
\(x=\tan \theta \\
dx=\sec^2 \theta d \theta\\

\int \dfrac{1}{\sec\theta }\sec^2 \theta d \theta=\int\sec \theta d \theta\\\)

xD

Answer 15

im feeling lazy to write but yeah we end up with
\(ln|\sec \theta+\tan \theta |+c=ln|\sqrt{1+x^2}+x|+c\)

Answer 16

why not use a u-sub?

Answer 17

you mean (arsinhx)' @Jhannybean

Answer 18

hmm that's quite something
how do we verify that that's arsinhx @marki

Answer 19

Yeah, \[d(\sinh x) \implies \frac{d}{dx} (\sinh x) \]

Answer 20

Actually, I was doing a hyperbolic u-sub, \(x = \sinh x\)

Answer 21

\[\int \frac{dx}{\sqrt{1+x^2}}\]Letting \(x=\sinh u~,~ dx = \cosh u~du\)\[\int \frac{\cosh u}{\sqrt{1+(\sinh u)^2}}\]\[1+(\sinh u)^2 = (\cosh u)^2\]\[\int \frac{\cosh u~du}{\cosh u}\]\[\int du = u+C = \sinh^{-1} (x) + C \]

Answer 22

can you make \(u=\sqrt{1+x^2}\) ?

Answer 23

not nearly as quick as what you did

Answer 24

\[\int \frac{dx}{\sqrt{1+x^2}}\]\[u= \sqrt{1+x^2} \]\[du = \frac{1}{2}(1+x^2)^{-1/2}\cdot 2x = \frac{x}{2(1+x^2)^{1/2}}\]

Answer 25

How would you replace the x in the numerator of \(du\)?

Answer 26

Oh, oops, typo.

Answer 27

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

Answer 28

probably not a good idea, but i think have seen it done with no trig
but maybe not \[u=\sqrt{1+x^2},x=\sqrt{u^2-1, }du=\frac{x}{\sqrt{1+x^2}}, du=\frac{\sqrt{u^2-1}}{u}\]

Answer 29

Oh it took me a minute to realize that. \[\color{red}{u=\sqrt{1+x^2}~,~ x =\sqrt{1-u^2}}\]\[du =\frac{x}{\sqrt{1+x^2}} \implies du = \frac{\sqrt{1-u^2}}{u} \] Haha

Answer 30

Oh wait, \[u^2=1+x^2 \implies x^2 = u^2-1 \implies x = \sqrt{u^2-1}\] My bad.

Answer 31

there's no edit button dangit.

Answer 32

thank @perl and move on

Answer 33

Lol of course.<.<

Answer 34

i meant it's \(D(sinh^{-1}x)=\Large \frac{1}{\sqrt{1+x^2}}\)
not D(sinhx)