1. Problem a) Find the integral $$\int \frac{6}{x^2 + 3} \, dx$$ Formula: $$\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan \frac{x}{a} + C$$ Here, $a^2 = 3$, so $a = \sqrt{3}$. Step 1: Factor out the constant 6: $$\int \frac{6}{x^2 + 3} \, dx = 6 \int \frac{1}{x^2 + (\sqrt{3})^2} \, dx$$ Step 2: Apply the formula: $$6 \cdot \frac{1}{\sqrt{3}} \arctan \frac{x}{\sqrt{3}} + C = \frac{6}{\sqrt{3}} \arctan \frac{x}{\sqrt{3}} + C$$ Step 3: Simplify the coefficient: $$\frac{6}{\sqrt{3}} = 2 \sqrt{3}$$ Final answer: $$\int \frac{6}{x^2 + 3} \, dx = 2 \sqrt{3} \arctan \frac{x}{\sqrt{3}} + C$$ 2. Problem b) Find the integral $$\int \sin \left( \frac{x}{2} \right) dx$$ Formula: $$\int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C$$ Here, $k = \frac{1}{2}$. Step 1: Apply the formula: $$\int \sin \left( \frac{x}{2} \right) dx = -\frac{1}{\frac{1}{2}} \cos \left( \frac{x}{2} \right) + C = -2 \cos \left( \frac{x}{2} \right) + C$$ 3. Problem c) Find the integral $$\int \frac{1}{2x - 1} \, dx$$ Formula: $$\int \frac{1}{ax + b} \, dx = \frac{1}{a} \ln |ax + b| + C$$ Here, $a=2$, $b=-1$. Step 1: Apply the formula: $$\int \frac{1}{2x - 1} \, dx = \frac{1}{2} \ln |2x - 1| + C$$ 4. Problem d) Find the integral $$\int \frac{x}{2 - x^2} \, dx$$ Rewrite denominator: $2 - x^2 = -(x^2 - 2)$. Step 1: Substitute $u = 2 - x^2$, then $du = -2x \, dx$ or $-\frac{1}{2} du = x \, dx$. Step 2: Rewrite integral: $$\int \frac{x}{2 - x^2} \, dx = \int \frac{x}{u} \, dx = \int \frac{-\frac{1}{2} du}{u} = -\frac{1}{2} \int \frac{1}{u} \, du$$ Step 3: Integrate: $$-\frac{1}{2} \ln |u| + C = -\frac{1}{2} \ln |2 - x^2| + C$$ 5. Problem e) Find the integral $$\int \frac{4x - 10}{x^2 - 5x + 1} \, dx$$ Step 1: Notice numerator is derivative of denominator: $$\frac{d}{dx} (x^2 - 5x + 1) = 2x - 5$$ Step 2: Rewrite numerator: $$4x - 10 = 2(2x - 5)$$ Step 3: Split integral: $$\int \frac{4x - 10}{x^2 - 5x + 1} \, dx = 2 \int \frac{2x - 5}{x^2 - 5x + 1} \, dx$$ Step 4: Use substitution $u = x^2 - 5x + 1$, $du = (2x - 5) dx$: $$2 \int \frac{du}{u} = 2 \ln |u| + C = 2 \ln |x^2 - 5x + 1| + C$$ 6. Problem f) Find the integral $$\int \tan 3x \, dx$$ Formula: $$\int \tan(kx) \, dx = -\frac{1}{k} \ln |\cos(kx)| + C$$ Here, $k=3$. Step 1: Apply formula: $$\int \tan 3x \, dx = -\frac{1}{3} \ln |\cos 3x| + C$$ Final answers: a) $$2 \sqrt{3} \arctan \frac{x}{\sqrt{3}} + C$$ b) $$-2 \cos \frac{x}{2} + C$$ c) $$\frac{1}{2} \ln |2x - 1| + C$$ d) $$-\frac{1}{2} \ln |2 - x^2| + C$$ e) $$2 \ln |x^2 - 5x + 1| + C$$ f) $$-\frac{1}{3} \ln |\cos 3x| + C$$