How to integrate x/(x⁴+1)

[tandoorichicken]

Is there a relatively simple way to solve this integral? Because I feel like I should know how to solve it, but I can't think of any way to do it.

$$\lim_{n\rightarrow\infty}\int_{1}^{n} \frac{x\,dx}{x^4+1}$$

 

[Data]

Factor the denominator of the integrand:

$$x^4 + 1 = (x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)$$

then use partial fractions to simplify and integrate. Don't worry about the limit until after that... your expression is really the same as

$$\int_1^\infty \frac{xdx}{x^4+1}$$

 

[Data]

Or on the other hand you could just sub $u = x^2$ without doing anything~

I'm silly!

 

[Jameson]

This is the arctangent rule.

Let u = x^2
du = 2xdx
du / 2 = xdx

$$\int_{1}^{n}\frac{xdx}{x^4+1} = \frac{1}{2}\int \frac{du}{u^2+1}$$

$$= \frac{\arctan{u}}{2} = \frac{\arctan{x^2}}{2}$$

Now apply the bounds and take the limit.

 

[p53ud0 dr34m5]

$$\lim_{n \to \infty} \int_1^n \frac{xdx}{x^4+1}$$
and you guys got this is equal to:
$$\lim_{n \to \infty}\frac{arctan(n^2)}{2}-\frac{arctan(1)}{2}=\lim_{n \to \infty}\frac{arctan(n^2)}{2}-0.3926990815=0.785398165 - 0.3926990815=0.3926990835$$

i think that's it. i don't have a book or anything, so i don't really know. hopefully that helps.

 

[dextercioby]

You needn't have put in decimal form (BTW,both $\frac{\pi}{4}$ and $\frac{\pi}{8}$ have an infinite # of decimals)...

So the integral is

$$\int_{1}^{+\infty} \frac{x}{x^{4}+1} \ dx =\frac{\pi}{8}$$

Daniel.

 

[p53ud0 dr34m5]

heh, thanks for clearing up what i did. all i had was this stupid computers calculator; I am at work. I am surprised i even got the right answer, haha.

 

[dextercioby]

Incidentally,your subtraction is wrong

Daniel.

 

[p53ud0 dr34m5]

arggg, its right. look at it!
$$0.785398165 - 0.3926990815 \approx 0.392699082 \approx \frac{pi}{8}$$
its legit

 

[dextercioby]

Check the last decimals and blame it on the lousy computer...

Daniel.