The given integral is:

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

To solve this, we'll try partial fraction decomposition or substitution if applicable, based on the structure of the denominator.

Step 1: Factor the denominator

We start by factoring the cubic polynomial $x^3 - x^2 - x + 1$, if possible.

Let's check if the cubic polynomial has any rational roots using the Rational Root Theorem.

The possible rational roots are factors of the constant term (1) over the factors of the leading coefficient (1), so the potential roots are $\pm 1$.

Testing $x = 1$: $1^3 - 1^2 - 1 + 1 = 1 - 1 - 1 + 1 = 0$ Thus, $x = 1$ is a root.

Step 2: Perform polynomial division

Since $x = 1$ is a root, we can divide $x^3 - x^2 - x + 1$ by $x - 1$.

Performing synthetic or long division of $x^3 - x^2 - x + 1$ by $x - 1$ yields:

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

Now, the integral becomes:

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

Step 3: Partial fraction decomposition

We express the fraction as:

$\frac{3x + 5}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$

Multiplying both sides by $(x - 1)(x^2 + 1)$, we get:

$3x + 5 = A(x^2 + 1) + (Bx + C)(x - 1)$

Expanding the right-hand side:

$3x + 5 = A(x^2 + 1) + (Bx + C)(x - 1)$ $= A(x^2 + 1) + Bx(x - 1) + C(x - 1)$ $= A(x^2 + 1) + Bx^2 - Bx + Cx - C$ $= A(x^2 + 1) + Bx^2 + (C - B)x - C$

Simplifying:

$3x + 5 = (A + B)x^2 + (C - B)x + (A - C)$

Now, we can compare coefficients on both sides of the equation:

• Coefficient of $x^2$: $A + B = 0$
• Coefficient of $x$: $C - B = 3$
• Constant term: $A - C = 5$

Step 4: Solve the system of equations

From $A + B = 0$, we get $B = -A$.

Substitute $B = -A$ into $C - B = 3$:

$C + A = 3$ $C = 3 - A$

Substitute $C = 3 - A$ into $A - C = 5$:

$A - (3 - A) = 5$ $A - 3 + A = 5$ $2A = 8 \quad \Rightarrow \quad A = 4$

Now, $B = -A = -4$ and $C = 3 - A = 3 - 4 = -1$.

Step 5: Substitute values into the partial fractions

We now have:

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

Step 6: Integrate each term

Now, integrate each term separately:

1. $\int \frac{4}{x - 1} \, dx = 4 \ln |x - 1|$

2. $\int \frac{-4x - 1}{x^2 + 1} \, dx$:

   • The term $\int \frac{-4x}{x^2 + 1} \, dx$ can be solved using the substitution $u = x^2 + 1$, yielding $-2 \ln(x^2 + 1)$.
   • The term $\int \frac{-1}{x^2 + 1} \, dx = -\tan^{-1}(x)$.

Final Answer:

Thus, the integral is:

$4 \ln |x - 1| - 2 \ln(x^2 + 1) - \tan^{-1}(x) + C$

Would you like more details or have any questions? Here are some related questions:

1. How is the Rational Root Theorem used in solving polynomial equations?
2. Can all cubic polynomials be factored, and if not, how are they handled?
3. What is the process of synthetic division, and how does it compare to long division?
4. How can partial fractions be applied to higher-order polynomials?
5. What are the key substitution methods for integrating rational functions?

Tip: Always try to factor the denominator of rational functions first, as it simplifies the integration process.