We are asked to evaluate three integrals: i. $\int 3x^2 + \sqrt{x} \, dx$ ii. $\int_{1}^{2} \frac{1}{x} - \frac{1}{x^2} \, dx$ iii. $\int e^{2x-3} \, dx$

1. Problem Description

We are asked to evaluate three integrals:

i. $\int 3x^2 + \sqrt{x} \, dx$

ii. $\int_{1}^{2} \frac{1}{x} - \frac{1}{x^2} \, dx$

iii. $\int e^{2x-3} \, dx$

2. Solution Steps

i.

We need to evaluate the indefinite integral $\int 3x^2 + \sqrt{x} \, dx$.

First, rewrite $\sqrt{x}$ as $x^{1/2}$.

$\int 3x^2 + x^{1/2} \, dx = \int 3x^2 \, dx + \int x^{1/2} \, dx$

Using the power rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, we have:

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

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

Therefore, $\int 3x^2 + \sqrt{x} \, dx = x^3 + \frac{2}{3}x^{3/2} + C$.

ii.

We need to evaluate the definite integral $\int_{1}^{2} \frac{1}{x} - \frac{1}{x^2} \, dx$.

$\int_{1}^{2} \frac{1}{x} - \frac{1}{x^2} \, dx = \int_{1}^{2} \frac{1}{x} \, dx - \int_{1}^{2} \frac{1}{x^2} \, dx$

We know that $\int \frac{1}{x} \, dx = \ln|x| + C$.

Also, $\int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = \frac{x^{-1}}{-1} = -\frac{1}{x} + C$.

Therefore, $\int_{1}^{2} \frac{1}{x} - \frac{1}{x^2} \, dx = [\ln|x| + \frac{1}{x}]_{1}^{2} = (\ln(2) + \frac{1}{2}) - (\ln(1) + \frac{1}{1})$

Since $\ln(1) = 0$, we have $\ln(2) + \frac{1}{2} - 1 = \ln(2) - \frac{1}{2}$.

iii.

We need to evaluate the indefinite integral $\int e^{2x-3} \, dx$.

Let $u = 2x - 3$. Then $\frac{du}{dx} = 2$, so $dx = \frac{1}{2}du$.

Then $\int e^{2x-3} \, dx = \int e^u \cdot \frac{1}{2} \, du = \frac{1}{2} \int e^u \, du = \frac{1}{2}e^u + C = \frac{1}{2}e^{2x-3} + C$.

3. Final Answer

i. $x^3 + \frac{2}{3}x^{3/2} + C$

ii. $\ln(2) - \frac{1}{2}$

iii. $\frac{1}{2}e^{2x-3} + C$