Question

use substitution techniques and a table of integrals to find the indefinite integral.

$$\int \frac { x } { \sqrt { x ^ { 4 } - 81 } } d x$$

click the icon to view a brief table of integrals.

choose the most useful substitution below.

a. $$u = x ^ { 4 }$$
b. $$u = x ^ { 4 } - 81$$
c. $$u = x$$
d. $$u = \sqrt { x ^ { 4 } - 81 }$$
e. $$u = x ^ { 2 }$$

Explanation:

Step1: Analyze substitution options

Let's check each option.

• If \(u = x^{4}\), then \(du=4x^{3}dx\), which is not directly helpful for the integrand \(\frac{x}{\sqrt{x^{4}-81}}dx\).
• If \(u = x^{4}-81\), then \(du = 4x^{3}dx\), not directly related to the integrand.
• If \(u=x\), then \(du = dx\), no simplification for \(\sqrt{x^{4}-81}\).
• If \(u=\sqrt{x^{4}-81}\), then \(du=\frac{4x^{3}}{2\sqrt{x^{4}-81}}dx=\frac{2x^{3}}{\sqrt{x^{4}-81}}dx\), not helpful.
• If \(u = x^{2}\), then \(du=2xdx\), and \(x^{4}=(x^{2})^{2}=u^{2}\). The integral \(\int\frac{x}{\sqrt{x^{4}-81}}dx=\frac{1}{2}\int\frac{du}{\sqrt{u^{2}-81}}\).

Answer:

E. \(u = x^{2}\)