Key Concepts and Formulas

• Integration by Substitution: If we can express an integral in the form $\int f(g(x))g'(x) \, dx$, then we can substitute $u = g(x)$, so $du = g'(x) \, dx$, and the integral becomes $\int f(u) \, du$.
• Logarithmic Differentiation/Integration: $\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C$.
• Properties of Logarithms: $\ln a - \ln b = \ln \frac{a}{b}$.

Step-by-Step Solution

Step 1: Factor the denominator. We start by factoring the denominator of the integrand: $\int \frac{2x^3 - 1}{x^4 + x} \, dx = \int \frac{2x^3 - 1}{x(x^3 + 1)} \, dx$

Step 2: Manipulate the integrand to apply substitution. Divide both numerator and denominator by $x^2$ to get the form $\frac{f'(x)}{f(x)}$: $\int \frac{2x^3 - 1}{x(x^3 + 1)} \, dx = \int \frac{2x - \frac{1}{x^2}}{x^2 + \frac{1}{x}} \, dx$

Step 3: Perform the substitution. Let $u = x^2 + \frac{1}{x} = x^2 + x^{-1}$. Then, $\frac{du}{dx} = 2x - \frac{1}{x^2}$. Thus, $du = (2x - \frac{1}{x^2}) \, dx$. The integral becomes: $\int \frac{2x - \frac{1}{x^2}}{x^2 + \frac{1}{x}} \, dx = \int \frac{du}{u}$

Step 4: Integrate with respect to $u$. $\int \frac{du}{u} = \ln |u| + C$

Step 5: Substitute back for $x$. Substitute $u = x^2 + \frac{1}{x}$ back into the expression: $\ln |u| + C = \ln \left|x^2 + \frac{1}{x}\right| + C = \ln \left|\frac{x^3 + 1}{x}\right| + C$

Step 6: Use logarithm properties to rewrite the expression. $\ln \left|\frac{x^3 + 1}{x}\right| + C = \ln |x^3 + 1| - \ln |x| + C$ We can rewrite this as: $\ln |x^3 + 1| - \ln |x| + C = \ln|x^3+1| + C' - \ln|x| + C''$ where $C = C' + C''$. This is equivalent to $\ln|x^3+1| - \ln|x| + C = \ln |x^3+1| - \ln |x| + \ln e^C = \ln\left| \frac{x^3+1}{x} \right| + C$

Step 7: Compare with the options. The expression $\ln \left|\frac{x^3 + 1}{x}\right| + C$ matches option (C). However, we need to arrive at option (A). Let's manipulate the expression further.

Step 8: Re-examine the given answer and try to reach it.

Given answer is $\ln_e \left| \frac{x^3 + 1}{x^2} \right| + C$. Our current result is $\ln \left|\frac{x^3 + 1}{x}\right| + C$. Somewhere we made a mistake. Let's go back to Step 2.

Step 2 (Revised): Instead of dividing by $x^2$, we should have kept $x$ in the denominator. $\int \frac{2x^3 - 1}{x(x^3 + 1)} \, dx = \int \frac{2x^3 - 1}{x^4 + x} \, dx$ Now, we want to write the numerator as a derivative of the denominator. Notice that the derivative of $x^4 + x$ is $4x^3 + 1$. Let's try to manipulate the integral to get this form: $\int \frac{2x^3 - 1}{x^4 + x} \, dx$ It doesn't seem like we can directly write the numerator as a constant times the derivative of the denominator.

Let's try partial fraction decomposition instead. $\frac{2x^3 - 1}{x(x^3 + 1)} = \frac{A}{x} + \frac{Bx^2 + Cx + D}{x^3 + 1}$ $2x^3 - 1 = A(x^3 + 1) + (Bx^2 + Cx + D)x = Ax^3 + A + Bx^3 + Cx^2 + Dx$ Comparing coefficients: $x^3: 2 = A + B$ $x^2: 0 = C$ $x: 0 = D$ Constant: $-1 = A$ So, $A = -1$, $B = 3$, $C = 0$, $D = 0$. $\int \frac{2x^3 - 1}{x(x^3 + 1)} \, dx = \int \left( \frac{-1}{x} + \frac{3x^2}{x^3 + 1} \right) \, dx = -\int \frac{1}{x} \, dx + 3 \int \frac{x^2}{x^3 + 1} \, dx$ $= -\ln |x| + \ln |x^3 + 1| + C = \ln |x^3 + 1| - \ln |x| + C = \ln \left| \frac{x^3 + 1}{x} \right| + C$ The problem states that the answer is $\ln_e \left| \frac{x^3 + 1}{x^2} \right| + C$.

Let's examine where we went wrong.

We want to get $\ln \left| \frac{x^3 + 1}{x^2} \right| + C$. This is equal to $\ln |x^3 + 1| - \ln |x^2| + C = \ln |x^3 + 1| - 2\ln |x| + C$.

The correct split should give us $\frac{2x^3 - 1}{x(x^3 + 1)} = \frac{A}{x} + \frac{Bx^2 + Cx + D}{x^3 + 1}$ $2x^3 - 1 = A(x^3 + 1) + x(Bx^2 + Cx + D)$ $2x^3 - 1 = Ax^3 + A + Bx^3 + Cx^2 + Dx$ $2 = A + B$, $0 = C$, $0 = D$, $-1 = A$. So $A = -1$, $B = 3$, $C = 0$, $D = 0$. $\int \frac{-1}{x} + \frac{3x^2}{x^3 + 1} dx = -\ln |x| + \ln |x^3 + 1| + C$ This is still $\ln \left| \frac{x^3 + 1}{x} \right| + C$.

Let's try to manipulate the equation to get the required answer. $\ln \left| \frac{x^3 + 1}{x} \right| = \ln \left| \frac{x^3 + 1}{x^2} \right| + \ln |x|$. This is not correct.

Let's check the given answer. $\ln \left| \frac{x^3 + 1}{x^2} \right| + C$. If we differentiate this, we get $\frac{x^2}{x^3 + 1} \cdot \frac{3x^2(x^2) - 2x(x^3 + 1)}{x^4} = \frac{3x^4 - 2x^4 - 2x}{x^2(x^3 + 1)} = \frac{x^4 - 2x}{x^2(x^3 + 1)} = \frac{x^3 - 2}{x(x^3 + 1)}$. This is not equal to $\frac{2x^3 - 1}{x(x^3 + 1)}$. So the answer is wrong.

The answer should be $\ln \left| \frac{x^3 + 1}{x} \right| + C$.

Common Mistakes & Tips

• Double-check the differentiation of the final answer to ensure it matches the original integrand.
• Partial fraction decomposition can be useful when substitution doesn't immediately work.
• Remember logarithm properties to simplify the final expression.

Summary

We factored the denominator, used partial fraction decomposition and integration, and simplified the expression using logarithm properties. The integral evaluates to $\ln \left| \frac{x^3 + 1}{x} \right| + C$. The given correct answer, however, is $\ln \left| \frac{x^3 + 1}{x^2} \right| + C$. Since the provided correct answer is incorrect, we will derive what it should be. The correct answer to the integral is $\ln \left| \frac{x^3 + 1}{x} \right| + C$.

Final Answer

The final answer is $\boxed{{\log _e}\left| {{{{x^3} + 1} \over x}} \right| + C}$. This corresponds to option (C).