Another double integral problem

[8614smith]

Homework Statement

sketch the region of integration, and evaluate the integral by choosing the best order of integration
$$\int^{8}_{0}\int^{2}_{x^{1/3}}\frac{dydx}{y^{4}+1}$$

Homework Equations

integration by parts

The Attempt at a Solution

after sketching the graph and changing the limits I've got to:
$$\int^{2}_{0}\int^{y^3}_{0}\frac{dydx}{y^{4}+1}$$

integrating:

$$\int^{2}_{0}\left[\frac{x}{y^{4}+1}\right]^{y^3}_{0}dy=\int^{2}_{0}\frac{y^3}{{y^4}+1}dy$$

$$u = {y^3}$$

$$u' = 3{y^2}$$

$$v = ln({y^4}+1)$$

$$v' = \frac{1}{{y^4}+1}$$

$$\left[{y^3}ln({y^4}+1)-\int3{y^2}ln({y^4}+1)\right]$$

$$u = ln({y^4}+1)$$

$$u' = \frac{1}{{y^4}+1}$$

$$v = {y^3}$$

$$v' = 3{y^2}$$

$$= {y^3}ln({y^4}+1)-\int{y^3}\frac{1}{{y^4}+1}$$

And this is where i get stuck as its just an infinite loop of integration by parts, any ideas?

 

[rock.freak667]

From here

$$\int^{2}_{0}\frac{y^3}{{y^4}+1}dy$$

Just sub u=y⁴+1EDIT:

Also

$$\frac{d}{dy}[ln(y^4+1)] \neq \frac{1}{y^4 +1}$$

 

[8614smith]

From here

$$\int^{2}_{0}\frac{y^3}{{y^4}+1}dy$$

Just sub u=y⁴+1


EDIT:

Also

$$\frac{d}{dy}[ln(y^4+1)] \neq \frac{1}{y^4 +1}$$

Ok I've done that and i got the right answer :) but what is $$\int\frac{1}{{y^4}+1}$$ if its not $$ln({y^4}+1)$$ ?

I remember how to integrate ln functions it using the joke:
Q: what's the integral of 1 over cabin?
A: log cabin

 

[Altabeh]

Ok I've done that and i got the right answer :) but what is $$\int\frac{1}{{y^4}+1}$$ if its not $$ln({y^4}+1)$$ ?

That integral is too hard to be calculated by hand. But I've prepared it for you to not make mistakes like that: $$1/8\,\sqrt {2}\ln \left( {\frac {{x}^{2}+x\sqrt {2}+1}{{x}^{2}-x \sqrt {2}+1}} \right) +1/4\,\sqrt {2}\arctan \left( x\sqrt {2}+1\right) +1/4\,\sqrt {2}\arctan \left( x\sqrt {2}-1 \right)$$.

AB

 

[8614smith]

That integral is too hard to be calculated by hand. But I've prepared it for you to not make mistakes like that: $$1/8\,\sqrt {2}\ln \left( {\frac {{x}^{2}+x\sqrt {2}+1}{{x}^{2}-x \sqrt {2}+1}} \right) +1/4\,\sqrt {2}\arctan \left( x\sqrt {2}+1\right) +1/4\,\sqrt {2}\arctan \left( x\sqrt {2}-1 \right)$$.

AB

I'll have to take your word on that one, let's just hope it doesn't come up in my exam!