Key Concepts and Formulas

• Indefinite Integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$ and $C$ is the constant of integration.
• Substitution Method: If $\int f(g(x))g'(x) dx$ exists, then substituting $u = g(x)$ and $du = g'(x) dx$ transforms the integral into $\int f(u) du$.
• Algebraic manipulation and simplification.

Step-by-Step Solution

Step 1: Rewrite the integral We are given the integral: $I = \int \frac{dx}{x^2(x^4 + 1)^{3/4}}$ Our goal is to simplify this integral using substitution.

Step 2: Manipulate the integrand We rewrite the integrand by factoring out $x^4$ from the term inside the parentheses: $I = \int \frac{dx}{x^2(x^4(1 + x^{-4}))^{3/4}} = \int \frac{dx}{x^2(x^4)^{3/4}(1 + x^{-4})^{3/4}} = \int \frac{dx}{x^2 \cdot x^3 (1 + x^{-4})^{3/4}} = \int \frac{dx}{x^5 (1 + x^{-4})^{3/4}}$

Step 3: Perform the substitution Let $y = 1 + x^{-4}$. Then, $\frac{dy}{dx} = -4x^{-5}$, which means $dy = -4x^{-5} dx$. Therefore, $x^{-5} dx = -\frac{1}{4} dy$. Substituting this into the integral, we get: $I = \int \frac{dx}{x^5 (1 + x^{-4})^{3/4}} = \int \frac{1}{(1 + x^{-4})^{3/4}} \cdot \frac{dx}{x^5} = \int \frac{1}{y^{3/4}} \cdot \left(-\frac{1}{4} dy\right) = -\frac{1}{4} \int y^{-3/4} dy$

Step 4: Evaluate the integral Now we can easily integrate with respect to $y$: $I = -\frac{1}{4} \int y^{-3/4} dy = -\frac{1}{4} \cdot \frac{y^{-3/4 + 1}}{-3/4 + 1} + C = -\frac{1}{4} \cdot \frac{y^{1/4}}{1/4} + C = -y^{1/4} + C$

Step 5: Substitute back for x Substituting $y = 1 + x^{-4}$ back into the expression, we have: $I = -(1 + x^{-4})^{1/4} + C = -\left(1 + \frac{1}{x^4}\right)^{1/4} + C = -\left(\frac{x^4 + 1}{x^4}\right)^{1/4} + C$

Step 6: Verify the answer matches the correct answer. The correct answer provided is $- {\left( {{x^4} + 1} \right)^{{1 \over 4}}} + c$. Our answer is $- \left(\frac{x^4 + 1}{x^4}\right)^{1/4} + C$. Let's differentiate the correct answer. $\frac{d}{dx} \left( - (x^4+1)^{1/4} + C \right) = - \frac{1}{4} (x^4+1)^{-3/4} (4x^3) = - \frac{x^3}{(x^4+1)^{3/4}}$ This is not the original integrand.

Let's differentiate our answer. $\frac{d}{dx} \left( - \left(\frac{x^4 + 1}{x^4}\right)^{1/4} + C \right) = - \frac{1}{4} \left(\frac{x^4 + 1}{x^4}\right)^{-3/4} \left(\frac{4x^3(x^4) - 4x^3(x^4+1)}{x^8} \right) = -\frac{1}{4} \left(\frac{x^4 + 1}{x^4}\right)^{-3/4} \left( \frac{4x^7 - 4x^7 - 4x^3}{x^8} \right) = -\frac{1}{4} \left(\frac{x^4 + 1}{x^4}\right)^{-3/4} \left( \frac{-4x^3}{x^8} \right) = \frac{x^{-5}}{\left( \frac{x^4+1}{x^4} \right)^{3/4}} = \frac{1}{x^5 \frac{(x^4+1)^{3/4}}{x^3}} = \frac{1}{x^2 (x^4+1)^{3/4}}$ This matches the original integrand.

The provided correct answer is incorrect.

Common Mistakes & Tips

• Be careful with algebraic manipulations, especially when dealing with fractional exponents.
• Remember to substitute back to the original variable after integration.
• Always check your answer by differentiating to see if you get back the original integrand.

Summary

We solved the given indefinite integral by first manipulating the integrand to isolate a suitable term for substitution. After performing the substitution and integrating, we substituted back to express the result in terms of the original variable, x. The final answer is $- {\left( {{{{x^4} + 1} \over {{x^4}}}} \right)^{{1 \over 4}}} + C$. The correct answer provided is incorrect.

Final Answer

The final answer is \boxed{-{\left( {{{{x^4} + 1} \over {{x^4}}}} \right)^{{1 \over 4}}} + c}, which corresponds to option (B).