Any simple methods of solving this?

Question

The question was initially: $$\int{\frac{1}{x^4+1}dx}$$ By trig substitution I reached the below equation and I'm stuck here: $$\int\frac{1}{2}{(\cot x)^{1/2}dx}$$
How can I proceed from here?Any simple methods are appreciated.

Answer 1

\begin{align} x^4 + 1 = \Big( x^4 + 2x^2 + 1\Big) - 2x^2 = {} & \Big( x^2+1\Big)^2 - \Big(x\sqrt2\Big)^2 \\ & \qquad \text{ This } \uparrow \text{ is a difference of two squares} \\ & \qquad \text{and can be factored accordingly.} \\[10pt] = {} & (x^2 + 1 -x\sqrt 2\,)(x^2+1 + x\sqrt 2\,) \end{align}

Each of these two quadratic factors is irreducible unless you bring in imaginary numbers, and there are various reasons for not doing that on this occasion. You can tell it's irreducible by looking at the discriminant $b^2-4ac$ where $a=1, b = \pm\sqrt 2, c=1,$ and observing that this discriminant is negative.

Then use partial fractions. You'll see something like this: $$ \int\frac{ax+b}{x^2 + 1 - x\sqrt 2} \, dx\quad \text{ plus another similar integral, and you need to find $a$ and $b$.} $$ Letting $u = x^2 +1-x\sqrt 2,$ you get $du = (2x - \sqrt 2\,)\,dx,$ and so $$ (ax+b)\,dx = \frac a 2 \left( 2x + \frac{2b} a \right) \, dx = \underbrace{\frac a 2 (2x - \sqrt2 \,)\,dx}{} + \left( b + \frac{a\sqrt2} 2 \right) \,dx $$ For the term over the $\underbrace{\text{underbrace}},$ just use the substitution.

For the other term, you'll have \begin{align} & \overbrace{\int \frac k {x^2 + 1 - x\sqrt 2} \, dx = \int \frac k {\left( x^2 - x\sqrt 2 + \frac 1 2 \right) + \frac 1 2} \, dx }^\text{completing the square} \\[12pt] = {} & \int \frac k {\left( x - \frac 1 {\sqrt 2} \right)^2 + \frac 1 2} \,dx = \int \frac{2k}{ \left( x\sqrt 2 - 1 \right)^2 + 1 } \, dx = \int \frac{2k}{w^2 + 1} \,\, \frac{dw}{\sqrt 2} = \cdots \end{align} and then use the fact that $\displaystyle \int \frac{dw}{w^2+1} = \arctan w + C.$