Question

The integral \[ \int \frac{dx}{x^2 \left( x^4 + 1 \right)^{3/4}} \] equals:

• A. \( \left( x^4 + 1 \right)^{1/4} + c \)
• B. \( - \left( x^4 + 1 \right)^{1/4} + c \)
• C. \( - \frac{(x^4 + 1)^{1/4}}{x^4} + c \)
• D. \( \left( \frac{x^4 + 1}{x^4} \right)^{1/4} + c \)

Hint:

Always check the limits of substitution and ensure they align with the integrand before performing integration by substitution.

Answer 1

The Correct Option is C

We are given the integral: \[ \int \frac{dx}{x^2 \left( x^4 + 1 \right)^{3/4}} \] To solve it, we make a substitution. Let \( 1 + \frac{1}{x^4} = t \), so that we can simplify the integral. Then, \[ -4x^5 \, dx = dt \quad \text{or} \quad x^5 \, dx = -\frac{dt}{4} \] Now, substitute into the integral: \[ \int \frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} \, dx \] Simplifies to: \[ - \frac{1}{4} \int \frac{dt}{x^5} \, \left( x^4 + 1 \right)^{3/4} \] We can now solve the integral and find: \[ - \frac{(x^4 + 1)^{1/4}}{x^4} + c \] Thus, the correct answer is \( - \frac{(x^4 + 1)^{1/4}}{x^4} + c \).