Evaluate the integrals in Exercises 1–14.
∫ 5 dx / √(25x² - 9), where x > 3/5

Verified step by step guidance

1. Recognize that the integral has the form \( \int \frac{5}{\sqrt{25x^2 - 9}} \, dx \), which resembles an integral involving \( \sqrt{a^2 x^2 - b^2} \). Here, identify \( a^2 = 25 \) and \( b^2 = 9 \), so \( a = 5 \) and \( b = 3 \).
2. Since the expression under the square root is \( 25x^2 - 9 = (5x)^2 - 3^2 \), consider using a hyperbolic substitution because \( x > \frac{3}{5} \) ensures the expression inside the square root is positive. Use the substitution \( 5x = 3 \cosh(t) \), which implies \( x = \frac{3}{5} \cosh(t) \).
3. Differentiate \( x = \frac{3}{5} \cosh(t) \) to find \( dx = \frac{3}{5} \sinh(t) \, dt \). Also, rewrite the square root in terms of \( t \): \( \sqrt{25x^2 - 9} = \sqrt{9 \cosh^2(t) - 9} = 3 \sinh(t) \).
4. Substitute all parts back into the integral: replace \( dx \) and the square root expression, and simplify the integral in terms of \( t \). This should reduce the integral to a simpler form involving hyperbolic functions.
5. Integrate with respect to \( t \), then substitute back \( t = \cosh^{-1}\left( \frac{5x}{3} \right) \) to express the answer in terms of \( x \).

Key Concepts

Integration of Rational Functions Involving Square Roots

This concept involves integrating functions where the integrand contains a rational expression with a square root in the denominator or numerator. Recognizing the form helps in choosing appropriate substitution or trigonometric identities to simplify the integral.

Trigonometric Substitution

Trigonometric substitution is a technique used to evaluate integrals involving expressions like √(a²x² - b²). By substituting x with a trigonometric function, the square root simplifies, allowing the integral to be expressed in terms of trigonometric functions that are easier to integrate.

Domain Restrictions and Absolute Values in Integration

When integrating functions with square roots, domain restrictions (like x > 3/5) ensure the expression under the root is positive. This affects the choice of substitution and the sign of the resulting expressions, especially when dealing with absolute values in antiderivatives.