Integrating 1/(u⁴+1), 1/(u⁵+1), and 1/(u⁶+1)

[DH]

Determine the integral:
y = \int_{0}^{1} 1/(u^4+1)du
and
y = \int_{0}^{1} 1/(u^5+1)du
and
y = \int_{0}^{1} 1/(u^6+1)du

 

[Data]

Just factor the denominator (to quadratics) and use partial fractions.

 

[DH]

Just factor the denominator (to quadratics) and use partial fractions.

no no, can't do like that!

 

[Data]

don't see why not~

 

[DH]

don't see why not~

show me the solution!

 

[Data]

\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)}

= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right)

= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)

and from there it's just minor substitutions to finish.

 

[DH]

Hix, you did the wrong thing form row 1 -> row 2!

 

[DH]

\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)}

= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right)

= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)

and from there it's just minor substitutions to finish.

how about this:
y = \int_{0}^{1} 1/(u^5+1)du

 

[Data]

Same advice. You can do it yourself this time

 

[DH]

Same advice. You can do it yourself this time

but one more time, can you show me the solution! I really want to know the way you solve it to compare with my own method... Please give me the solution in detail!

 

[Data]

You can do it almost exactly the same way I did the last one - factor the denominator (in this case, it factors to a product of two irreducible quadractics and the linear factor (x+1)), then use partial fractions to separate it into a sum of functions that you know how to integrate.

Why don't you post an example of a solution using your method? I'm interested now!

 

[DH]

\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)}

= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right)

= \frac{1}{2\sqrt{2}} \left( \int_0^1 \frac{du}{\left(u-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} - \int_0^1 \frac{du}{\left(u+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}} \right)

and from there it's just minor substitutions to finish.

You made a mistake here:
\int_0^1 \frac{du}{u^4+1} = \int_0^1 \frac{du}{(u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)}

= \int_0^1 \frac{du}{2\sqrt{2}}\left(\frac{1}{u^2 - \sqrt{2}u + 1} - \frac{1}{u^2 + \sqrt{2}u + 1} \right)

 

[Data]

indeed I did, the second line should be

\int_0^1 \frac{du}{2}\left( \frac{\frac{1}{\sqrt{2}}u + 1}{u^2+\sqrt{2}u+1} - \frac{\frac{1}{\sqrt{2}}u - 1}{u^2-\sqrt{2}u+1}\right),

but the simplification after that is still similar, just with more terms.