Question

Can do you help me with a integration with trigonometric sustitution? https://s.yimg.com/hd/answers/i/8f396c265d4a4d3185e0ce08f5849d38_A.png?a=answers&mr=0&x=1446167478&s=de2c02fa472f497b20e4bb5eafd58b33

Answer 1

It's hard? :(

Answer 2

The only thing I can see is if you let
$$
y=\cfrac{\sqrt{\sqrt{1+x^4}-x^2}}{x}
$$
Then
$$
\cfrac{dy}{dx}=\cfrac{1}{x^2\sqrt{x^4+1}\sqrt{\sqrt{1+x^4}-x^2}}
$$
Which is close to what you have in your integrand. If we multiply this whole thing by
$$
\cfrac{x^2}{\sqrt{x^4+1}}
$$
We would have your integrad
But then this equals
$$
\cfrac{1}{1-y^2}
$$
So
$$
\cfrac{1}{1-y^2}\cfrac{dy}{dx}=\cfrac{1}{(1+x^4)\sqrt{\sqrt{1+x^4}-x^2}}
$$
Does this help?

Answer 3

Notice that
$$
\cfrac{d}{dx}\coth^{-1}y=\cfrac{1}{1-y^2}\cfrac{dy}{dx}
$$
https://en.wikipedia.org/wiki/Inverse_hyperbolic_function#Derivatives

Answer 4

I think your answer is correct and thank you for your answer :D. How could this be done with trigonometric sustitution? Because that was my teacher requested, and I don't know too much about this :(

Answer 5

your link is unopenable, at this end at least

Answer 6

i have no idea