Question

Integral of x/sqrt (x^2+4)

Answer 1

try \(u=x^2+4\)

Answer 2

it's for trigonometric sustitution...

Answer 3

lol why?

Answer 4

you can do it in your head
\[u=x^2+4, du =2xdx, \frac{1}{2}du=dx\]
\[\frac{1}{2}\int \frac{du}{\sqrt{u}}=\sqrt{u}=\sqrt{x^2+4}\]

Answer 5

we can do it with a trig sub if you like
put \(x=\tan(\theta), dx=\sec^2(\theta)d\theta\)

Answer 6

oops, forgot about the 4
put \[x=2\tan(\theta), dx=2\sec^2(\theta)d\theta\]

Answer 7

good so far?

Answer 8

yes, i did that, but i don't know what to do next \[\int\limits2\tan \theta /\sqrt{2\tan^2\theta+4}\]

Answer 9

you forgot the \(2\sec^2(\theta)\) up top

Answer 10

also you forgot to square the 2 inside the radical

Answer 11

oh yes, i forgot that, anyway what should i do next?

Answer 12

\[\int \frac{2\tan(\theta)2\sec^2(\theta)}{\sqrt{4\tan^2(\theta)+4}}d\theta\]

Answer 13

then \(\sqrt{4\tan^2(\theta)+4}=\sqrt{4(\tan^2(\theta)+1)}=2\sec(\theta)\)

Answer 14

cancel a 2
cancel a secant, get
\[2\int \sec(\theta)\tan(\theta)d\theta\]

Answer 15

a function whose anti derivative you know right?

Answer 16

yes, thanks i appreciate it :)

Answer 17

this is really a long chasing the tail method of doing it
you are converting to trig and converting back, but no matter, if this is what you have to do, then you can do it

Answer 18

yw