We are asked to solve two integration problems. First, we need to evaluate $\int (2\sqrt{x} + 5x^{-1} + 3x^2) dx$. Second, we need to evaluate $\int (3^x + e^x + x^{-7/2}) dx$.

1. Problem Description

We are asked to solve two integration problems.

First, we need to evaluate $\int (2\sqrt{x} + 5x^{-1} + 3x^2) dx$.

Second, we need to evaluate $\int (3^x + e^x + x^{-7/2}) dx$.

2. Solution Steps

First integral: $\int (2\sqrt{x} + 5x^{-1} + 3x^2) dx$

We can rewrite the integral as the sum of three integrals:

$\int 2\sqrt{x} dx + \int 5x^{-1} dx + \int 3x^2 dx$

$\int 2x^{1/2} dx + \int \frac{5}{x} dx + \int 3x^2 dx$

We can use the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, and $\int \frac{1}{x} dx = \ln|x| + C$.

$2\int x^{1/2} dx = 2 \cdot \frac{x^{3/2}}{3/2} = 2 \cdot \frac{2}{3} x^{3/2} = \frac{4}{3}x^{3/2}$

$5\int \frac{1}{x} dx = 5\ln|x|$

$3\int x^2 dx = 3 \cdot \frac{x^3}{3} = x^3$

So the first integral becomes $\frac{4}{3}x^{3/2} + 5\ln|x| + x^3 + C$

Second integral: $\int (3^x + e^x + x^{-7/2}) dx$

Again, we can rewrite the integral as the sum of three integrals:

$\int 3^x dx + \int e^x dx + \int x^{-7/2} dx$

We know that $\int a^x dx = \frac{a^x}{\ln a} + C$, $\int e^x dx = e^x + C$, and we can use the power rule for the last integral.

$\int 3^x dx = \frac{3^x}{\ln 3} + C$

$\int e^x dx = e^x + C$

$\int x^{-7/2} dx = \frac{x^{-5/2}}{-5/2} + C = -\frac{2}{5}x^{-5/2} + C$

So the second integral becomes $\frac{3^x}{\ln 3} + e^x - \frac{2}{5}x^{-5/2} + C$

3. Final Answer

The first integral is $\frac{4}{3}x^{3/2} + 5\ln|x| + x^3 + C$.

The second integral is $\frac{3^x}{\ln 3} + e^x - \frac{2}{5}x^{-5/2} + C$.