Question

∫2xdx/sqrt(x^4+16)

Answer 1

http://www.wolframalpha.com/input/?i=2x%2Fsqrt%28x^4%2B16%29+dx

Answer 2

it has the detailed solution

Answer 3

my webwork says its wrong

Answer 4

because you did not cover inverse hyperbolic sines that is why

Answer 5

can you solve this without hyperbolic sinfunction, i havent been taught that yet

Answer 6

maybe
\[u=x^2\]
\[du=2xdx\] and get
\[\int \frac{du}{\sqrt{u^2+16}}\]

Answer 7

do you mind showing me th rest?

Answer 8

i think this works. you use a trig sub here
\[z=4\tan(u)\]
\[dz = 4\sec^2(u)du\]

Answer 9

and so you get
\[\int\sec(z)dz\] after getting everything out of the radical

Answer 10

then you look in your book for the anti - derivative of secant and see that it is
\[\ln(\tan(z)+\sec(z))\]

Answer 11

\[Ln(\frac{x^2+ \sqrt{x^4+16}}{4} )+ C\]

Answer 12

it says +C is not defined in this context

Answer 13

and since we started with
\[z=4\tan(u)\] you get
\[u=\tan^{-1}(\frac{z}{4})\]so at we have
\[\ln(\frac{u}{4}+\frac{\sqrt{u^2+16}}{4})\]

Answer 14

and finally if i did not screw this up too much, we started with
\[u=x^2\] so our "final answer " is
\[\ln(\frac{x^2}{4}+\frac{\sqrt{x^4+16}}{4})\]

Answer 15

i need the final answer now before its too late, then you can explain

Answer 16

which looks good because it is what valpey wrote, so i am happy, although i was sure i messed up along the way. god i hate these

Answer 17

We can check with :
\[\frac{d(ln(\frac{x^2}{4}+\frac{\sqrt{x^4+16}}{4}))}{dx} = \frac{\frac{d(\frac{x^2}{4}+\frac{\sqrt{x^4+16}}{4})}{dx}}{\frac{x^2}{4}+\frac{\sqrt{x^4+16}}{4}}\]

Answer 18

\[=\frac{\frac{x}{2}+\frac{x^3(x^4+16)^{-1/2}}{2}}{\frac{x^2}{4}+\frac{\sqrt{x^4+16}}{4}} = \frac{2x+\frac{2x^3}{\sqrt{x^4+16}}}{x^2 + \sqrt{x^4+16}}\]
\[= \frac{2x\sqrt{x^4+16} + 2x^3}{x^2\sqrt{x^4+16}+x^4+16}\]
\[=
\frac{2x(\sqrt{x^4+16} + x^2)}{\sqrt{x^4+16}(\sqrt{x^4+16}+x^2)} = \frac{2x}{\sqrt{x^4+16}}\]

Answer 19

Wow, this editor is really bogged down. I'm going to try writing my script in a different program.