To C or not to C - split -Integral

[Steven G]

Consider the standard integral of [imath]\arcsin(u)[/imath], which is
[math]\arcsin(u)=\int \frac{1}{\sqrt{1-u^2}}\,du[/math]Notice your problem resembles it (in red).
[math]\int\frac{x}{\sqrt{3-x^4}}\,dx=\int\frac{x}{\sqrt{3}\sqrt{1-\frac{x^4}{3}}}\,dx=\int\frac{x}{\sqrt{3}}\cdot \red{\frac{1}{\sqrt{1-\frac{x^4}{3}}}}\,dx[/math]If you recognized that, it's an obvious choice to make a u-subsitution [imath]u^2=\frac{x^4}{3}[/imath] or [imath]u=\frac{x^2}{\sqrt{3}}[/imath].
I'll let you work out the details but clearly, this isn't something you're expected to do and I second Dr.P's comments. It's more valuable for you.

+C

 

[Deleted member 4993]

+C

Where do you want put that '+C'

The response in response #11 has the integral sign - the process of integration has not been finished.

But the corner is really dark - so you may not have noticed it.

 

[Steven G]

Where do you want put that '+C'

The response in response #11 has the integral sign - the process of integration has not been finished.

But the corner is really dark - so you may not have noticed it.

The 2nd line of the post should have a +C. To the corner SM.

 

[Deleted member 4993]

arcsin(u) → ∫1/(1−u²)du

This one?? I would agree to put a 'C' if it were written the other way:

∫1/(1−u²)du = arcsin(u) + C

Do we also have to state that 'u' is a real number?

(and thank you for speaking-up for me in another thread.)

 

[topsquark]

The 2nd line of the post should have a +C. To the corner SM.

"Knowledge is a light in the darkness of the corner." - Quote by probably someone

-Dan

 

[Steven G]

(and thank you for speaking-up for me in another thread.)

You're my boss and I will always defend you.