The problem asks us to evaluate two integrals and two limits. (a) $\int \frac{x^2}{2} dx$ (b) $\int \frac{x}{x+1} dx$ (c) $\lim_{x \to 3} \frac{1}{x-3}$ (d) $\lim_{x \to 0} \frac{2x+2}{3x+5}$

1. Problem Description

The problem asks us to evaluate two integrals and two limits.

(a) $\int \frac{x^2}{2} dx$

(b) $\int \frac{x}{x+1} dx$

(c) $\lim_{x \to 3} \frac{1}{x-3}$

(d) $\lim_{x \to 0} \frac{2x+2}{3x+5}$

2. Solution Steps

(a) Integral of $x^2/2$

$\int \frac{x^2}{2} dx = \frac{1}{2} \int x^2 dx$

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

$\frac{1}{2} \int x^2 dx = \frac{1}{2} \frac{x^3}{3} + C = \frac{x^3}{6} + C$

(b) Integral of $x/(x+1)$

$\int \frac{x}{x+1} dx$

We can rewrite the integrand as follows:

$\frac{x}{x+1} = \frac{x+1-1}{x+1} = \frac{x+1}{x+1} - \frac{1}{x+1} = 1 - \frac{1}{x+1}$

So, the integral becomes:

$\int (1 - \frac{1}{x+1}) dx = \int 1 dx - \int \frac{1}{x+1} dx = x - \ln|x+1| + C$

(c) Limit of $1/(x-3)$ as $x$ approaches

3. $\lim_{x \to 3} \frac{1}{x-3}$

As $x$ approaches 3 from the left (i.e., $x \to 3^-$), $x-3$ approaches 0 from the negative side. Therefore, $\frac{1}{x-3}$ approaches $-\infty$.

As $x$ approaches 3 from the right (i.e., $x \to 3^+$), $x-3$ approaches 0 from the positive side. Therefore, $\frac{1}{x-3}$ approaches $+\infty$.

Since the left and right limits are not equal, the limit does not exist.

(d) Limit of $(2x+2)/(3x+5)$ as $x$ approaches

0. $\lim_{x \to 0} \frac{2x+2}{3x+5}$

We can evaluate the limit by substituting $x=0$ into the expression since the function is continuous at $x=0$.

$\frac{2(0)+2}{3(0)+5} = \frac{2}{5}$

3. Final Answer

(a) $\int \frac{x^2}{2} dx = \frac{x^3}{6} + C$

(b) $\int \frac{x}{x+1} dx = x - \ln|x+1| + C$

(c) $\lim_{x \to 3} \frac{1}{x-3}$ does not exist.

(d) $\lim_{x \to 0} \frac{2x+2}{3x+5} = \frac{2}{5}$