Question

\( \int \frac{1}{x^2 (x^4 + 1)^{3/4 dx \) is equal to:}

• A. \( -\frac{(1 + x^4)^{3/4}}{x} + C \)
• B. \( -\frac{(1 + x^4)^{1/4}}{2x} + C \)
• C. \( -\frac{(1 + x^4)^{1/4}}{x} + C \)
• D. \( -\frac{(1 + x^4)^{1/4}}{x^2} + C \)
• E. \( -\frac{(1 + x^4)^{1/2}}{x} + C \)

Hint:

For integrals like this, "taking out" the highest power of $x$ from the root/fractional power is a standard trick to create the derivative of the inner function outside.

Answer 1

The Correct Option is C

Concept: We use the substitution method by factoring out the highest power of \( x \) from the expression inside the fractional power.

Step 1: Rearrange the integrand.
\[ \int \frac{1}{x^2 (x^4(1 + x^{-4}))^{3/4}} dx = \int \frac{1}{x^2 \cdot x^3 (1 + x^{-4})^{3/4}} dx = \int \frac{1}{x^5 (1 + x^{-4})^{3/4}} dx \]

Step 2: Use substitution.
Let \( u = 1 + x^{-4} \). Then \( du = -4x^{-5} dx \), which implies \( \frac{1}{x^5} dx = -\frac{1}{4} du \). \[ \int \left(-\frac{1}{4}\right) \frac{du}{u^{3/4}} = -\frac{1}{4} \int u^{-3/4} du \]

Step 3: Integrate and back-substitute.
\[ -\frac{1}{4} \left( \frac{u^{1/4}}{1/4} \right) + C = -u^{1/4} + C \] \[ = -(1 + x^{-4})^{1/4} + C = -\left( \frac{x^4 + 1}{x^4} \right)^{1/4} + C = -\frac{(1 + x^4)^{1/4}}{x} + C \]