The image contains several mathematics problems. We will solve problem number 8 which asks us to find the integral $J = \int \frac{6x^2 + 4x}{x^3 + x^2 - x - 1} dx$.

1. Problem Description

The image contains several mathematics problems. We will solve problem number 8 which asks us to find the integral $J = \int \frac{6x^2 + 4x}{x^3 + x^2 - x - 1} dx$.

2. Solution Steps

First, factor the denominator.

$x^3 + x^2 - x - 1 = x^2(x+1) - (x+1) = (x^2 - 1)(x+1) = (x-1)(x+1)(x+1) = (x-1)(x+1)^2$.

Then we can rewrite the integral as:

$J = \int \frac{6x^2 + 4x}{(x-1)(x+1)^2} dx$.

We can use partial fraction decomposition to split the fraction into simpler terms:

$\frac{6x^2 + 4x}{(x-1)(x+1)^2} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$

$6x^2 + 4x = A(x+1)^2 + B(x-1)(x+1) + C(x-1)$

$6x^2 + 4x = A(x^2 + 2x + 1) + B(x^2 - 1) + C(x-1)$

$6x^2 + 4x = Ax^2 + 2Ax + A + Bx^2 - B + Cx - C$

$6x^2 + 4x = (A+B)x^2 + (2A+C)x + (A-B-C)$

Comparing coefficients:

$A+B = 6$

$2A+C = 4$

$A-B-C = 0$

From the first equation, we have $B = 6-A$.

From the third equation, we have $C = A-B = A-(6-A) = 2A-6$.

Substituting $C$ into the second equation:

$2A + 2A - 6 = 4$

$4A = 10$

$A = \frac{10}{4} = \frac{5}{2}$.

$B = 6 - \frac{5}{2} = \frac{12}{2} - \frac{5}{2} = \frac{7}{2}$.

$C = 2(\frac{5}{2}) - 6 = 5-6 = -1$.

Thus, we have

$\frac{6x^2 + 4x}{(x-1)(x+1)^2} = \frac{5/2}{x-1} + \frac{7/2}{x+1} + \frac{-1}{(x+1)^2}$.

Now we can rewrite the integral:

$J = \int \left( \frac{5/2}{x-1} + \frac{7/2}{x+1} - \frac{1}{(x+1)^2} \right) dx$

$J = \frac{5}{2} \int \frac{1}{x-1} dx + \frac{7}{2} \int \frac{1}{x+1} dx - \int \frac{1}{(x+1)^2} dx$

$J = \frac{5}{2} \ln|x-1| + \frac{7}{2} \ln|x+1| + \frac{1}{x+1} + C$

3. Final Answer

$J = \frac{5}{2} \ln|x-1| + \frac{7}{2} \ln|x+1| + \frac{1}{x+1} + C$