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Evaluate the integral and express your answer in simplest form. int(-2)/(xsqrt(x^(2)-9))dx Answer:

Evaluate the integral and express your answer in simplest form.\newline∫−2xx2−9dx \int \frac{-2}{x \sqrt{x^{2}-9}} d x \newlineAnswer:

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Q. Evaluate the integral and express your answer in simplest form.\newline∫−2xx2−9dx \int \frac{-2}{x \sqrt{x^{2}-9}} d x \newlineAnswer:
  1. Identify Substitution: Let's identify a substitution that can simplify the integral. We notice that the denominator has the form of a derivative of a hyperbolic function, specifically the derivative of cosh⁡(x)\cosh(x), which is sinh⁡(x)\sinh(x). If we let x=3cosh⁡(u)x = 3\cosh(u), then dx=3sinh⁡(u)dudx = 3\sinh(u) du, and the integral becomes easier to handle. Let's perform this substitution.
  2. Perform Substitution: Substitute x=3cosh⁡(u)x = 3\cosh(u), which implies dx=3sinh⁡(u)dudx = 3\sinh(u) du. The integral becomes:\newline∫−23cosh⁡(u)9cosh⁡2(u)−9⋅3sinh⁡(u)du\int\frac{-2}{3\cosh(u)\sqrt{9\cosh^2(u)-9}} \cdot 3\sinh(u) du\newlineSimplify the square root:\newline∫−23cosh⁡(u)9(cosh⁡2(u)−1)⋅3sinh⁡(u)du\int\frac{-2}{3\cosh(u)\sqrt{9(\cosh^2(u)-1)}} \cdot 3\sinh(u) du\newlineSince cosh⁡2(u)−1=sinh⁡2(u)\cosh^2(u) - 1 = \sinh^2(u), we have:\newline∫−23cosh⁡(u)9sinh⁡2(u)⋅3sinh⁡(u)du\int\frac{-2}{3\cosh(u)\sqrt{9\sinh^2(u)}} \cdot 3\sinh(u) du
  3. Simplify Square Root: Simplify the integral further by canceling terms:\newline∫−23cosh⁡(u)⋅3sinh⁡(u)⋅3sinh⁡(u) du\int\frac{-2}{3\cosh(u)\cdot 3\sinh(u)} \cdot 3\sinh(u) \, du\newlineThe 3sinh⁡(u)3\sinh(u) terms cancel out, leaving us with:\newline∫−23cosh⁡(u) du\int\frac{-2}{3\cosh(u)} \, du
  4. Cancel Terms: Now, the integral simplifies to:\newline∫(−23)⋅(1cosh⁡(u)) du\int(-\frac{2}{3}) \cdot (\frac{1}{\cosh(u)}) \, du\newlineRecognize that 1cosh⁡(u)\frac{1}{\cosh(u)} is the hyperbolic secant function, \sech(u):\newline\int(-\frac{2}{3}) \cdot \sech(u) \, du
  5. Recognize Hyperbolic Secant: The integral of sech(u)\text{sech}(u) is a standard result, which is −23ln⁡∣tanh⁡(u2)∣+C-\frac{2}{3} \ln|\tanh(\frac{u}{2})| + C, where CC is the constant of integration. So we have:\newline−23ln⁡∣tanh⁡(u2)∣+C-\frac{2}{3} \ln|\tanh(\frac{u}{2})| + C
  6. Back-Substitute Variables: Now we need to back-substitute to return to the original variable xx. We have x=3cosh⁡(u)x = 3\cosh(u), so we need to express tanh⁡(u2)\tanh(\frac{u}{2}) in terms of xx. We can use the half-angle formula for hyperbolic tangent:\newlinetanh⁡(u2)=cosh⁡(u)−1sinh⁡(u)\tanh(\frac{u}{2}) = \frac{\cosh(u) - 1}{\sinh(u)}\newlineSince x=3cosh⁡(u)x = 3\cosh(u), we have cosh⁡(u)=x3\cosh(u) = \frac{x}{3} and sinh⁡(u)=x2−93\sinh(u) = \frac{\sqrt{x^2 - 9}}{3}. Substituting these into the half-angle formula gives us:\newlinetanh⁡(u2)=(x3)−1x2−93\tanh(\frac{u}{2}) = \frac{(\frac{x}{3}) - 1}{\frac{\sqrt{x^2 - 9}}{3}}
  7. Substitute Back into Integral: Substitute tanh⁡(u2)\tanh(\frac{u}{2}) back into the integral result:\newline−23⋅ln⁡∣(x3)−1x2−93∣+C-\frac{2}{3} \cdot \ln\left|\frac{(\frac{x}{3}) - 1}{\frac{\sqrt{x^2 - 9}}{3}}\right| + C\newlineSimplify the expression:\newline−23⋅ln⁡∣x−3x2−9∣+C-\frac{2}{3} \cdot \ln\left|\frac{x - 3}{\sqrt{x^2 - 9}}\right| + C
  8. Simplify Expression: The final answer in simplest form is: −23ln⁡∣(x−3)x2−9∣+C-\frac{2}{3} \ln\left|\frac{(x - 3)}{\sqrt{x^2 - 9}}\right| + C

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