1. We are asked to find the integrals: $$I_1 = \int 2xe^x \, dx, \quad I_2 = \int 2x \cos x \, dx, \quad I_3 = \int 2x \sin x \, dx, \quad I_4 = \int 2x \ln x \, dx$$ 2. We will use integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ where we choose $u$ and $dv$ appropriately. --- ### For $I_1 = \int 2xe^x \, dx$: 1. Let $u = 2x$, so $du = 2 \, dx$. 2. Let $dv = e^x \, dx$, so $v = e^x$. 3. Apply integration by parts: $$I_1 = uv - \int v \, du = 2x e^x - \int e^x \cdot 2 \, dx = 2x e^x - 2 \int e^x \, dx$$ 4. Evaluate the remaining integral: $$\int e^x \, dx = e^x + C$$ 5. So, $$I_1 = 2x e^x - 2 e^x + C = 2 e^x (x - 1) + C$$ --- ### For $I_2 = \int 2x \cos x \, dx$: 1. Let $u = 2x$, so $du = 2 \, dx$. 2. Let $dv = \cos x \, dx$, so $v = \sin x$. 3. Apply integration by parts: $$I_2 = uv - \int v \, du = 2x \sin x - \int \sin x \cdot 2 \, dx = 2x \sin x - 2 \int \sin x \, dx$$ 4. Evaluate the remaining integral: $$\int \sin x \, dx = -\cos x + C$$ 5. So, $$I_2 = 2x \sin x - 2 (-\cos x) + C = 2x \sin x + 2 \cos x + C$$ --- ### For $I_3 = \int 2x \sin x \, dx$: 1. Let $u = 2x$, so $du = 2 \, dx$. 2. Let $dv = \sin x \, dx$, so $v = -\cos x$. 3. Apply integration by parts: $$I_3 = uv - \int v \, du = 2x (-\cos x) - \int (-\cos x) \cdot 2 \, dx = -2x \cos x + 2 \int \cos x \, dx$$ 4. Evaluate the remaining integral: $$\int \cos x \, dx = \sin x + C$$ 5. So, $$I_3 = -2x \cos x + 2 \sin x + C$$ --- ### For $I_4 = \int 2x \ln x \, dx$: 1. Let $u = \ln x$, so $du = \frac{1}{x} \, dx$. 2. Let $dv = 2x \, dx$, so $v = \int 2x \, dx = x^2$. 3. Apply integration by parts: $$I_4 = uv - \int v \, du = x^2 \ln x - \int x^2 \cdot \frac{1}{x} \, dx = x^2 \ln x - \int x \, dx$$ 4. Evaluate the remaining integral: $$\int x \, dx = \frac{x^2}{2} + C$$ 5. So, $$I_4 = x^2 \ln x - \frac{x^2}{2} + C$$ --- Now for the second set: $$I_1 = \int (3x^2 + 4x + 1) \cos x \, dx, \quad I_2 = \int (3x^2 + 4x + 1) \ln x \, dx$$ --- ### For $I_1 = \int (3x^2 + 4x + 1) \cos x \, dx$: We use integration by parts multiple times or treat as sum of integrals: $$I_1 = \int 3x^2 \cos x \, dx + \int 4x \cos x \, dx + \int \cos x \, dx$$ We solve each separately. 1. For $\int 3x^2 \cos x \, dx$: Let $u = 3x^2$, $du = 6x \, dx$. Let $dv = \cos x \, dx$, $v = \sin x$. Apply parts: $$= 3x^2 \sin x - \int 6x \sin x \, dx$$ 2. For $\int 6x \sin x \, dx$: Let $u = 6x$, $du = 6 \, dx$. Let $dv = \sin x \, dx$, $v = -\cos x$. Apply parts: $$= -6x \cos x + \int 6 \cos x \, dx = -6x \cos x + 6 \sin x + C$$ So, $$\int 3x^2 \cos x \, dx = 3x^2 \sin x - (-6x \cos x + 6 \sin x) + C = 3x^2 \sin x + 6x \cos x - 6 \sin x + C$$ 3. For $\int 4x \cos x \, dx$: Let $u = 4x$, $du = 4 \, dx$. Let $dv = \cos x \, dx$, $v = \sin x$. Apply parts: $$= 4x \sin x - \int 4 \sin x \, dx = 4x \sin x + 4 \cos x + C$$ 4. For $\int \cos x \, dx = \sin x + C$ 5. Combine all: $$I_1 = (3x^2 \sin x + 6x \cos x - 6 \sin x) + (4x \sin x + 4 \cos x) + \sin x + C$$ Simplify: $$I_1 = 3x^2 \sin x + 6x \cos x - 6 \sin x + 4x \sin x + 4 \cos x + \sin x + C$$ $$= 3x^2 \sin x + 4x \sin x + 6x \cos x + ( -6 \sin x + \sin x) + 4 \cos x + C$$ $$= 3x^2 \sin x + 4x \sin x + 6x \cos x - 5 \sin x + 4 \cos x + C$$ --- ### For $I_2 = \int (3x^2 + 4x + 1) \ln x \, dx$: Split integral: $$I_2 = \int 3x^2 \ln x \, dx + \int 4x \ln x \, dx + \int \ln x \, dx$$ Use integration by parts for each: 1. For $\int 3x^2 \ln x \, dx$: Let $u = \ln x$, $du = \frac{1}{x} \, dx$. Let $dv = 3x^2 \, dx$, $v = x^3$. Apply parts: $$= x^3 \ln x - \int x^3 \cdot \frac{1}{x} \, dx = x^3 \ln x - \int x^2 \, dx = x^3 \ln x - \frac{x^3}{3} + C$$ 2. For $\int 4x \ln x \, dx$: Let $u = \ln x$, $du = \frac{1}{x} \, dx$. Let $dv = 4x \, dx$, $v = 2x^2$. Apply parts: $$= 2x^2 \ln x - \int 2x^2 \cdot \frac{1}{x} \, dx = 2x^2 \ln x - \int 2x \, dx = 2x^2 \ln x - x^2 + C$$ 3. For $\int \ln x \, dx$: Use parts: Let $u = \ln x$, $du = \frac{1}{x} \, dx$. Let $dv = dx$, $v = x$. Apply parts: $$= x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int 1 \, dx = x \ln x - x + C$$ 4. Combine all: $$I_2 = (x^3 \ln x - \frac{x^3}{3}) + (2x^2 \ln x - x^2) + (x \ln x - x) + C$$ Simplify: $$I_2 = x^3 \ln x + 2x^2 \ln x + x \ln x - \frac{x^3}{3} - x^2 - x + C$$