1. **Problem Statement:** Evaluate the integrals: a) $$\int \frac{5}{4 - \sqrt{3} - z} \, dz$$ b) $$\int \frac{w}{w - 2\sqrt{6} - 3w - 5} \, dw$$ c) $$\int \cos(\sqrt{x}) \, dx$$ 2. **Integral a:** $$\int \frac{5}{4 - \sqrt{3} - z} \, dz$$ - This is a rational function integral of the form $$\int \frac{c}{a - z} \, dz$$ where $$c=5$$ and $$a=4 - \sqrt{3}$$. - The integral formula is $$\int \frac{1}{x - a} \, dx = \ln|x - a| + C$$. - Applying the constant multiple rule: $$\int \frac{5}{4 - \sqrt{3} - z} \, dz = -5 \int \frac{1}{z - (4 - \sqrt{3})} \, dz = -5 \ln|z - (4 - \sqrt{3})| + C$$ - Note the negative sign because denominator is $$4 - \sqrt{3} - z = -(z - (4 - \sqrt{3}))$$. 3. **Integral b:** $$\int \frac{w}{w - 2\sqrt{6} - 3w - 5} \, dw$$ - Simplify denominator: $$w - 2\sqrt{6} - 3w - 5 = -2w - 2\sqrt{6} - 5$$ - Rewrite integral: $$\int \frac{w}{-2w - 2\sqrt{6} - 5} \, dw = -\int \frac{w}{2w + 2\sqrt{6} + 5} \, dw$$ - Use substitution: $$u = 2w + 2\sqrt{6} + 5$$, so $$du = 2 dw$$, $$dw = \frac{du}{2}$$. - Express numerator in terms of $$u$$: $$w = \frac{u - 2\sqrt{6} - 5}{2}$$ - Substitute: $$-\int \frac{\frac{u - 2\sqrt{6} - 5}{2}}{u} \cdot \frac{du}{2} = -\int \frac{u - 2\sqrt{6} - 5}{4u} \, du = -\frac{1}{4} \int \left(1 - \frac{2\sqrt{6} + 5}{u}\right) du$$ - Integrate: $$-\frac{1}{4} \left( u - (2\sqrt{6} + 5) \ln|u| \right) + C$$ - Substitute back $$u$$: $$-\frac{1}{4} \left( 2w + 2\sqrt{6} + 5 - (2\sqrt{6} + 5) \ln|2w + 2\sqrt{6} + 5| \right) + C$$ 4. **Integral c:** $$\int \cos(\sqrt{x}) \, dx$$ - Use substitution: $$t = \sqrt{x} = x^{1/2}$$, so $$x = t^2$$, $$dx = 2t \, dt$$. - Rewrite integral: $$\int \cos(t) \, dx = \int \cos(t) \cdot 2t \, dt = 2 \int t \cos(t) \, dt$$ - Use integration by parts: Let $$u = t$$, $$dv = \cos(t) dt$$ Then $$du = dt$$, $$v = \sin(t)$$ So, $$2 \int t \cos(t) dt = 2 (t \sin(t) - \int \sin(t) dt) = 2 (t \sin(t) + \cos(t)) + C$$ - Substitute back $$t = \sqrt{x}$$: $$2 \sqrt{x} \sin(\sqrt{x}) + 2 \cos(\sqrt{x}) + C$$ **Final answers:** $$\int \frac{5}{4 - \sqrt{3} - z} \, dz = -5 \ln|z - (4 - \sqrt{3})| + C$$ $$\int \frac{w}{w - 2\sqrt{6} - 3w - 5} \, dw = -\frac{1}{4} \left( 2w + 2\sqrt{6} + 5 - (2\sqrt{6} + 5) \ln|2w + 2\sqrt{6} + 5| \right) + C$$ $$\int \cos(\sqrt{x}) \, dx = 2 \sqrt{x} \sin(\sqrt{x}) + 2 \cos(\sqrt{x}) + C$$