Integral on the real line between 0 and infinity using contour integration Question:

Let $f(z)=\frac{1}{1+z^{4}}$.
(a) Find the singularity of $f(z)$ in the first quadrant (where $\operatorname{Re} z, \operatorname{Im} z \geq 0$ ).
(b) Calculate the residue of $f(z)$ at the singular point lying in the first quadrant.
(c) Let $\Gamma_{R}$ be the quarter-circle $\Gamma_{R}:|z|=R, \operatorname{Re} z, \operatorname{Im} z \geq 0$, positively oriented. Show that $\lim _{R \rightarrow \infty} \int_{\Gamma_{R}} f(z) d z=0 .$
(d) Determine $\int_{0}^{\infty} \frac{1}{1+x^{4}} d x$.

For part (a) I have that the singularity is at $(1+i)/\sqrt2$ and it is a simple pole?
For part (b) I have that the residue at $f(z)$ at that point is $-(1+i)/4\sqrt2$
For part (c) I used the ML lemma (I am not very good using code so I do not know how to write out my solution without it looking very messy on here.
For part (d) I am a bit stuck, as I have only ever seen examples with the integral between $-\infty$ and $\infty$. Because it is an even function, is the integral between $0$ and $\infty$ just half of the integral between $-\infty$ and $\infty$? What do I do about the integral on the imaginary line?
I am confused... any help would be appreciated.

$x^4 = (ix)^4$, so the integral on the half imaginary axis is the same as the integral on the half real line.
