1. **Problem statement:** Find the integrals $I_1 = \int 2x e^x \, dx$ and $I_4 = \int 2x \ln x \, dx$.\n\n2. **Formula and rules:** For integrals involving products, use integration by parts: \n$$\int u \, dv = uv - \int v \, du$$\nChoose $u$ and $dv$ wisely to simplify the integral.\n\n3. **Integral $I_1 = \int 2x e^x \, dx$:**\n- Let $u = 2x$ so $du = 2 \, dx$.\n- Let $dv = e^x \, dx$ so $v = e^x$.\n\nApply integration by parts:\n$$I_1 = uv - \int v \, du = 2x e^x - \int e^x \cdot 2 \, dx = 2x e^x - 2 \int e^x \, dx$$\n\nIntegral of $e^x$ is $e^x$, so:\n$$I_1 = 2x e^x - 2 e^x + C = 2 e^x (x - 1) + C$$\n\n4. **Integral $I_4 = \int 2x \ln x \, dx$:**\n- Let $u = \ln x$ so $du = \frac{1}{x} dx$.\n- Let $dv = 2x \, dx$ so $v = \int 2x \, dx = x^2$.\n\nApply integration by parts:\n$$I_4 = uv - \int v \, du = x^2 \ln x - \int x^2 \cdot \frac{1}{x} \, dx = x^2 \ln x - \int x \, dx$$\n\nIntegral of $x$ is $\frac{x^2}{2}$, so:\n$$I_4 = x^2 \ln x - \frac{x^2}{2} + C$$\n\n**Final answers:**\n$$I_1 = 2 e^x (x - 1) + C$$\n$$I_4 = x^2 \ln x - \frac{x^2}{2} + C$$