\( \int 2 ^ { x } ( 1 + \ln x ) d x \)

To evaluate the integral: \[ \int 2^x (1 + \ln x) \, dx \] we can use **integration by parts**. Let's set: - \( u = 1 + \ln x \) - \( dv = 2^x \, dx \) First, compute the derivatives and integrals needed: \[ du = \frac{1}{x} \, dx \] \[ v = \int 2^x \, dx = \frac{2^x}{\ln 2} \] Applying the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Substituting the values: \[ \int 2^x (1 + \ln x) \, dx = \left(1 + \ln x\right) \frac{2^x}{\ln 2} - \int \frac{2^x}{\ln 2} \cdot \frac{1}{x} \, dx \] Simplifying the expression: \[ = \frac{(1 + \ln x) 2^x}{\ln 2} - \frac{1}{\ln 2} \int \frac{2^x}{x} \, dx \] The remaining integral, \(\int \frac{2^x}{x} \, dx\), is known as the **Exponential Integral** and is denoted by \( \text{Ei}(x \ln 2) \). Therefore, the final expression for the integral is: \[ \int 2^x (1 + \ln x) \, dx = \frac{(1 + \ln x) 2^x - \text{Ei}(x \ln 2)}{\ln 2} + C \] where \( C \) is the constant of integration. **Final Answer:** After expanding, an antiderivative is [(1 + ln x) 2ˣ – Ei(x ln 2)] divided by ln 2, plus C. Thus, ∫2^x(1+ln x)dx = [(1+ln x)2^x − Ei(x ln2)] ⁄ ln2 + C