We are asked to evaluate three indefinite integrals: a) $\int \frac{x^2}{2} dx$ b) $\int \frac{x}{x+1} dx$ c) $\int \frac{1}{(x+\frac{2}{3})(x-\frac{4}{5})} dx$

1. Problem Description

We are asked to evaluate three indefinite integrals:

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

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

c) $\int \frac{1}{(x+\frac{2}{3})(x-\frac{4}{5})} dx$

2. Solution Steps

a) We need to evaluate $\int \frac{x^2}{2} dx$.

We can rewrite the integral as $\frac{1}{2} \int x^2 dx$.

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

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

b) We need to evaluate $\int \frac{x}{x+1} dx$.

We can rewrite the integrand as $\frac{x}{x+1} = \frac{x+1-1}{x+1} = \frac{x+1}{x+1} - \frac{1}{x+1} = 1 - \frac{1}{x+1}$.

Now, we can integrate:

$\int \frac{x}{x+1} dx = \int (1 - \frac{1}{x+1}) dx = \int 1 dx - \int \frac{1}{x+1} dx$.

$\int 1 dx = x + C_1$.

$\int \frac{1}{x+1} dx = \ln |x+1| + C_2$.

Therefore, $\int \frac{x}{x+1} dx = x - \ln |x+1| + C$, where $C = C_1 - C_2$.

c) We need to evaluate $\int \frac{1}{(x+\frac{2}{3})(x-\frac{4}{5})} dx$.

We will use partial fraction decomposition. We want to find constants $A$ and $B$ such that:

$\frac{1}{(x+\frac{2}{3})(x-\frac{4}{5})} = \frac{A}{x+\frac{2}{3}} + \frac{B}{x-\frac{4}{5}}$.

Multiplying both sides by $(x+\frac{2}{3})(x-\frac{4}{5})$, we get:

$1 = A(x-\frac{4}{5}) + B(x+\frac{2}{3})$.

$1 = Ax - \frac{4}{5}A + Bx + \frac{2}{3}B$.

Equating coefficients of $x$: $A+B=0$, so $B=-A$.

Equating constant terms: $-\frac{4}{5}A + \frac{2}{3}B = 1$.

Substituting $B=-A$, we get $-\frac{4}{5}A - \frac{2}{3}A = 1$.

$-\frac{12}{15}A - \frac{10}{15}A = 1$.

$-\frac{22}{15}A = 1$, so $A = -\frac{15}{22}$.

$B = -A = \frac{15}{22}$.

So we have $\int \frac{1}{(x+\frac{2}{3})(x-\frac{4}{5})} dx = \int (\frac{-\frac{15}{22}}{x+\frac{2}{3}} + \frac{\frac{15}{22}}{x-\frac{4}{5}}) dx$.

$= -\frac{15}{22} \int \frac{1}{x+\frac{2}{3}} dx + \frac{15}{22} \int \frac{1}{x-\frac{4}{5}} dx$.

$= -\frac{15}{22} \ln |x+\frac{2}{3}| + \frac{15}{22} \ln |x-\frac{4}{5}| + C$.

$= \frac{15}{22} (\ln |x-\frac{4}{5}| - \ln |x+\frac{2}{3}|) + C$.

$= \frac{15}{22} \ln |\frac{x-\frac{4}{5}}{x+\frac{2}{3}}| + C$.

3. Final Answer

a) $\frac{x^3}{6} + C$

b) $x - \ln |x+1| + C$

c) $\frac{15}{22} \ln |\frac{x-\frac{4}{5}}{x+\frac{2}{3}}| + C$