Question

Evaluate the integrals in Exercises 1–14.∫ 5 dx / √(25x² - 9), where x \u003e 3/5

Accepted Answer

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 $$ b = 3 . $$Since the expression under the square root is $$ 25x^2 - 9 = (5x)^2 - 3^2 $$, consider using a hyperbolic substitution because $$ x \u003e \frac{3}{5} $$ ensures the expression inside the square root is positive. Use the substitution $$ 5x = 3 \cosh(t) $$, which implies $$ x = \frac{3}{5} \cosh(t) . $$Differentiate $$ x = \frac{3}{5} \cosh(t)  to $$find $$ dx = \frac{3}{5} \sinh(t) \, dt . $$Also, rewrite the square root in terms of $$ t $$: $$ \sqrt{25x^2 - 9} = \sqrt{9 \cosh^2(t) - 9} = 3 \sinh(t) . $$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. Integrate with respect to $$ t $$, then substitute back $$ t = \cosh^{-1}\left( \frac{5x}{3} \right)  to $$express the answer in terms of $$ x .$$

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

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Key Concepts

Integration of Rational Functions Involving Square Roots

Trigonometric Substitution

Domain Restrictions and Absolute Values in Integration

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