Step-by-Step Solution

(a) Evaluate: $\int_0^{2\pi} \frac{1}{1 + e^{\sin x}} dx $

Let $I = \int_0^{2\pi} \frac{1}{1 + e^{\sin x}} dx$

Using the property $\int_0^{a} f(x) dx = \int_0^{a} f(a-x) dx$, we have:

$I = \int_0^{2\pi} \frac{1}{1 + e^{\sin (2\pi - x)}} dx = \int_0^{2\pi} \frac{1}{1 + e^{-\sin x}} dx = \int_0^{2\pi} \frac{e^{\sin x}}{e^{\sin x} + 1} dx$

Adding the two expressions for $I$, we get:

$2I = \int_0^{2\pi} \frac{1}{1 + e^{\sin x}} dx + \int_0^{2\pi} \frac{e^{\sin x}}{1 + e^{\sin x}} dx = \int_0^{2\pi} \frac{1 + e^{\sin x}}{1 + e^{\sin x}} dx = \int_0^{2\pi} 1 dx$

$2I = [x]_0^{2\pi} = 2\pi$

$I = \pi$

(b) Find: $\int \frac{x⁴} { ((x-1)(x²+1))}dx.$

We need to perform polynomial long division first to simplify the integrand.

$\frac{x^4}{(x-1)(x^2+1)} = \frac{x^4}{x^3 - x^2 + x - 1}$

Performing long division, we get:

$x^4 = (x+1)(x^3 - x^2 + x - 1) + (2x^2 - 2x + 1)$

So, $\frac{x^4}{x^3 - x^2 + x - 1} = x + 1 + \frac{2x^2 - 2x + 1}{x^3 - x^2 + x - 1} = x + 1 + \frac{2x^2 - 2x + 1}{(x-1)(x^2+1)}$

Now, we perform partial fraction decomposition on the remaining fraction:

$\frac{2x^2 - 2x + 1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx + C}{x^2+1}$

$2x^2 - 2x + 1 = A(x^2+1) + (Bx+C)(x-1)$

Let $x = 1$: $2(1)^2 - 2(1) + 1 = A(1^2+1) \implies 1 = 2A \implies A = \frac{1}{2}$

$2x^2 - 2x + 1 = \frac{1}{2}(x^2+1) + (Bx+C)(x-1)$

$2x^2 - 2x + 1 = \frac{1}{2}x^2 + \frac{1}{2} + Bx^2 - Bx + Cx - C$

$2x^2 - 2x + 1 = (\frac{1}{2} + B)x^2 + (-B+C)x + (\frac{1}{2} - C)$

Comparing coefficients:

$x^2: 2 = \frac{1}{2} + B \implies B = \frac{3}{2}$

$x: -2 = -B + C \implies -2 = -\frac{3}{2} + C \implies C = -\frac{1}{2}$

So, $\frac{2x^2 - 2x + 1}{(x-1)(x^2+1)} = \frac{1/2}{x-1} + \frac{(3/2)x - 1/2}{x^2+1} = \frac{1}{2(x-1)} + \frac{3x-1}{2(x^2+1)}$

Now, we integrate:

$\int \frac{x^4}{(x-1)(x^2+1)} dx = \int (x + 1 + \frac{1}{2(x-1)} + \frac{3x-1}{2(x^2+1)}) dx$

$= \int x dx + \int 1 dx + \frac{1}{2} \int \frac{1}{x-1} dx + \frac{3}{2} \int \frac{x}{x^2+1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx$

$= \frac{x^2}{2} + x + \frac{1}{2} \ln|x-1| + \frac{3}{4} \ln(x^2+1) - \frac{1}{2} \arctan(x) + C$