Evaluate the integral using residue

evaluate the integral using residue
1. ∞
∫ 1/(1+x^4) dx
-∞

2. ∞
∫ x^2 / (1+x^2)^2 dx
0

3. ∞
∫ 1/(1+x^2)^4 dx
-∞

4. ∞
∫ cos(x) / [(x^2+pi^2)(x^2+pi^2/4)] dx
0

Since you're asking for these all at once, the simple version is as follows:
We use a sufficiently large semicircular contour C (in upper half plane) to contain all singularities of the integrand which lie in the upper half plane.

As long as the integrand f(z) is rational and sufficiently well-behaved (degree of denom. is at least 2 more than that of the numerator), we have
∫(-∞ to ∞) f(x) dx = 2πi * (sum of residues of f(z) in the upper half plane).
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1) f(z) = 1/(z^4 + 1) has simple poles when z^4 + 1 = 0.
==> z^4 = -1 = e^(πi)
==> z = e^(πi(1 + 2k)/4) for k = 0, 1, 2, 3.

Of these poles, only those for k = 0 and 1 are in the upper half plane.

Compute the residues:
lim(z→e^(πi(1 + 2k)/4)) (z - e^(πi(1 + 2k)/4)) * 1/(z^4 + 1)
= lim(z→e^(πi(1 + 2k)/4)) 1/(4z^3), via L'Hopital's Rule
= lim(z→e^(πi(1 + 2k)/4)) z/(4z^4)
= lim(z→e^(πi(1 + 2k)/4)) (-1/4) z, since z^4 = -1
= (-1/4) e^(πi(1 + 2k)/4)).

k = 0 ==> (-1/4) e^(πi/4) = (-1/4)(√2/2 + i√2/2)
k = 1 ==> (-1/4) e^(3πi/4) = (-1/4)(-√2/2 + i√2/2)

Hence, ∫(-∞ to ∞) dx/(1 + x^4)
= 2πi [(-1/4)(√2/2 + i√2/2) + (-1/4)(-√2/2 + i√2/2)]
= π√2/2.
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2) The only singularity of the integrand in the upper half plane is z = i, a double pole.

It has residue
lim(z→i) (d/dz) [(z - i)^2 * z^2/(1 + z^2)^2]
= lim(z→i) (d/dz) z^2/(z + i)^2
= lim(z→i) [2z(z + i)^2 - z^2 * 2(z + i)] / (z + i)^4
= -i/4.

Hence, ∫(-∞ to ∞) x^2 dx/(1 + x^2)^2 = 2πi * (-i/4) = π/2
==> ∫(0 to ∞) x^2 dx/(1 + x^2)^2 = π/4.
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3) Similar to 2; the only singularity of the integrand in the upper half plane is z = i,
this time a quadruple pole.