Question

\( \int \frac{\sqrt{5 + x^2}}{x^4} \, dx \) is equal to:

• A. \( \frac{1}{15} \left( 1 + \frac{5}{x^2} \right)^{3/2} + C \)
• B. \( \frac{-1}{15} \left( 1 + \frac{1}{x^2} \right)^{3/2} + C \)
• C. \( \frac{-1}{15} \left( 1 + \frac{5}{x^2} \right)^{3/2} + C \)
• D. \( \frac{1}{15} \left( 1 + \frac{1}{x^2} \right)^{3/2} + C \)
• E. \( \frac{-1}{10} \left( 1 + \frac{1}{x^2} \right)^{3/2} + C \)

Hint:

"Taking out $x$" from a square root changes it to $x$. This trick often makes the rest of the denominator look like the derivative of the inside of the root.

Answer 1

The Correct Option is C

Concept: For integrals with a higher power of \( x \) in the denominator, factor out the highest power of \( x \) from the radical to set up a \( u \)-substitution.

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

Step 2: Apply substitution.
Let \( u = 1 + 5x^{-2} \). Then \( du = -10x^{-3} \, dx \), which means \( \frac{1}{x^3} \, dx = -\frac{1}{10} \, du \). The integral becomes: \[ -\frac{1}{10} \int \sqrt{u} \, du = -\frac{1}{10} \int u^{1/2} \, du \]

Step 3: Integrate and back-substitute.
\[ -\frac{1}{10} \left( \frac{u^{3/2}}{3/2} \right) + C = -\frac{1}{10} \cdot \frac{2}{3} u^{3/2} + C = -\frac{1}{15} u^{3/2} + C \] Substituting \( u = 1 + \frac{5}{x^2} \): \[ -\frac{1}{15} \left( 1 + \frac{5}{x^2} \right)^{3/2} + C \]