Question

Integration:
\[\int {\frac{{dx}}
{{{x^4} + 1}}} \]

Answer 1

I have done to this
\[\frac{1}
{{{x^4} + 1}} = \frac{{\frac{x}
{{2\sqrt 2 }} + \frac{1}
{2}}}
{{{x^2} + \sqrt 2 x + 1}} + \frac{{\frac{{ - x}}
{{2\sqrt 2 }} + \frac{1}
{2}}}
{{{x^2} - \sqrt 2 x + 1}}\]

Answer 2

And I don't know how to do next. ....

Answer 3

what about this pls help t=3n+6n

Answer 4

t=9n lolz

Answer 5

hmm Okay I have this divide numerator and denominator by \(x^2\)

Answer 6

\[\int \frac{\frac{1}{x^2}}{x^2+ \frac{1}{x^2} +2 -2}\]

Answer 7

pls help need answer t=6-3n .....11 years old don't understand homework lol

Answer 8

Note that\[ x^4 + 1 = (x^2 + √2 x + 1)(x^2 -√2x + 1).\]
Now break \[\frac{1}{[ (x^2 + √2 x + 1)(x^2 -√2x + 1)]}\]
into partial fractions and integrate.