antiderivative of (1+x^4)^(1/2)?

Question

Need help with explanation and detail

Answer 1

lizabeth's is totally wrong. You can't just ignore the √ sign. And the constant of integration is by no means necessarily 1.

Integ √(1+x^4) dx

These are the standard trigonometric substitutions to get rid of √a²±x² or √x²±a²
http://www.sosmath.com/calculus/integration/trigsub/trigsub.html


Here is the solution after 7 wrong tries, seems like I was going in the wrong direction, we want the square on the x² side of the substitution not the RHS:
x² = tan t
Then √(1+x^4) = √(1+tan²t) = sec t
2x dx = sec²t dt
2tan t dx = sec²t dt
dx = sec²t/2tan t dt

So we transformed Integ √(1+x^4) dx to
Integ sec t * sec²t/2tan t dt
= Integ sec³t/2tan t dt
= Integ sect(1+tan²t)/2tan t dt
= 1/2 Integ [ sec t cot t + sect tan t dt ]
= 1/2 Integ [ csc t + sect tan t dt ]

Integ csc t dt is a standard form (ln |sec t + tan t| ), and we can spot
[ sect tan t ] = d/dt sec t

Integ = 1/2 Integ [ csc t + sect tan t dt ]
= 1/2 ( sec t + ln |sec t + tan t| ) + C

SOLUTION!! finally after 8 tries
___________________________________

Here are all the unsuccessful tries
(I left this in so you can learn from this by seeing what leads to a dead end, this is pretty educational):


So for √a²+x² , with a =1:
Let x = tan t, => 1+x² = sec²t ,
and dx = sec²t dt = (1+x²) dt
i.e. dt = cos²t dx

Then: 1+x^4 = (1+x²)² - 2x² = (1+x²)² - 2(1+x²) + 2
= (sec²t)² -2sec²t + 2
= (sec²t - 1)² + 1

Thus √(1+x^4) = √((sec²t - 1)² + 1)

So we transformed Integ √(1+x^4) dx to
Integ √((sec²t - 1)² + 1) * sec²t dt
which has the form √z² + 1²

Looks like you have to make a double-substitution now...
... for the second substitution you could try y = 1*sec s


Or maybe we do both substitutions in one by doing
either tan² t or sec t tan t . (Substituting sec(tan(t)) is a disaster.)

Let's try y = sec t tan t = sin t/ cos² t
dy = [cos² t * cos t - sin t *2cos t sin t] / cos^4 t dt
dy = [cos³ t - 2cos t sin² t] / cos^4 t dt
dy = [cos² t - 2 (1-cos² t)] / cos³ t dt
dy = (3cos² t - 2) / cos³ t dt

Again we want to transform: √(1+x^4) = √((sec²t - 1)² + 1)
to only contain terms in (sec t tan t)...

Or we could just go for the jugular with
x = √tan t
Then √(1+x^4) = √(1+tan²t) = sec t
And dx = 1/2√tan t * sec²t dt
dx = sec²t / 2√tan t dt
dx = (1+ tan²t) / 2√tan t dt
dx = 1/2 [ 1/√tan t + √tan t ] dt

So we transformed Integ √(1+x^4) dx to
1/2 Integ [ sec t * ( 1/√tan t + √tan t ) ] dt
not sure that's too promising...

Another outside alternative to remove the radical:
√(1+x^4) = √(1+x^4)*√(1-x^4) / √(1-x^4)
= √(1-x^8) / √(1-x^4)

Answer 2

antiderivative of (1 + x^4) ^ 1/2
= (1 + x^4) ^ (1/2 +1) / (1/2 +1)
= (1 + x^4) ^ (3/2) /(3/2)
= (2/3) (1 + x^4) ^ (3/2)

hope this hope
just remember to always add 1 when taking the anitderivtive.