Finding Complex Antiderivatives | Guidance for Tricky Functions

[OmniNewton]

Homework Statement

How would one go about finding the antiderivative to this function?

Homework Equations

N/A

The Attempt at a Solution

This problem has been rather tricky I have tried several attempts at the solution. My one solution consists of me factoring out the x^4. Looking for some guidance please.

Thank you!

 

[Mark44]

Homework Statement

How would one go about finding the antiderivative to this function?

Homework Equations

N/A

The Attempt at a Solution

This problem has been rather tricky I have tried several attempts at the solution. My one solution consists of me factoring out the x^4. Looking for some guidance please.

Thank you!

I would start with an ordinary substitution, u = x², and would complete the square in the radical. From there, a trig substitution seems promising.

 

[OmniNewton]

I would start with an ordinary substitution, u = x², and would complete the square in the radical. From there, a trig substitution seems promising.

OK I will work it out now thank you for the suggestion.

 

[OmniNewton]

OK I gave your suggestion an attempt.
I've arrived at the following after the substitution from x ---> u completing the square----> and back to x.

(x^4-1)
x^2((x^2+1/4)^2+ (3/4)))^(1/2)

 

[mfb]

Where is the point in substituting back before integration?
u=x^2+1/2 (not 1/4) was my first idea as well, but then you still have an ugly sqrt(u) in the denominator.

 

[OmniNewton]

if you let u= x^2+(1/2) I also have a hard time figuring out how to remove the dx and convert it to du if du= 2x

 

[mfb]

That's where the sqrt(u-1/2) comes in. Forgot the 1/2 in the previous post.
Hmm, u=x^2 is an easier substitution. The initial square root goes away anyway.

Thinking about it... standard partial fraction decomposition should work, with imaginary numbers to have the zero.

 

[OmniNewton]

That's where the sqrt(u-1/2) comes in. Forgot the 1/2 in the previous post.
Hmm, u=x^2 is an easier substitution. The initial square root goes away anyway.

Thinking about it... standard partial fraction decomposition should work.

Should i apply partial fraction before or after u substitution. Also sorry for still not getting it but when I u substitute I'm still having a hard time figuring out how to change the integral from being with respect to x to respect to u.

 

[mfb]

Well, if u=x², then x=+-sqrt(u) and du = 2x dx.

Partial fraction decomposition would be without substitution.