This integral involves a rational function and a square root in the denominator. The given expression is:

$\int \left( \frac{1}{\sqrt{(x - 1)(x - 5)}} + \frac{2x - 3}{x^2 - 4x + 13} \right) dx$

Steps to Solve:

1. Separate the integral: $\int \frac{1}{\sqrt{(x - 1)(x - 5)}} dx + \int \frac{2x - 3}{x^2 - 4x + 13} dx$

2. Simplify the first term:

   • The denominator $\sqrt{(x - 1)(x - 5)}$ suggests a substitution like $x = 3 + 2\cos\theta$, since it represents a square root of a quadratic in factored form.
   • Rewrite and substitute accordingly to simplify.

3. Simplify the second term:

   • The denominator $x^2 - 4x + 13$ is a quadratic. Complete the square: $x^2 - 4x + 13 = (x - 2)^2 + 9$
   • Use the substitution $u = x - 2$, so the integral becomes: $\int \frac{2u}{u^2 + 9} du$
   • Solve using standard forms for arctangent integrals.

Let me calculate and simplify these integrals step by step. One moment! sympy as sp

Define the variables

x = sp.symbols('x')

Define the two terms of the integral

term1 = 1 / sp.sqrt((x - 1) * (x - 5)) term2 = (2x - 3) / (x**2 - 4x + 13)

Complete the square for the second term denominator

denom_term2 = x**2 - 4*x + 13 completed_square = sp.simplify(denom_term2)

Integrate both terms

integral_term1 = sp.integrate(term1, x) integral_term2 = sp.integrate(term2, x)

integral_term1, integral_term2The solution involves the following steps for the two terms of the integral:

1. First term: The integral remains unevaluated in its current form due to the complexity: $\int \frac{1}{\sqrt{(x - 1)(x - 5)}} dx$ This may require a substitution like $x = 3 + 2\cos\theta$, which simplifies the square root. Additional steps would be needed for explicit evaluation.

2. Second term: After completing the square for the denominator $x^2 - 4x + 13 = (x - 2)^2 + 9$, the integral simplifies to: $\int \frac{2x - 3}{x^2 - 4x + 13} dx = \ln(x^2 - 4x + 13) + \frac{1}{3} \arctan\left(\frac{x - 2}{3}\right) + C$

Would you like me to explicitly evaluate the first term further or explain the substitution process? Let me know!

Relative Questions:

1. How do substitutions simplify square roots in integrals?
2. Can trigonometric substitutions always be applied to simplify expressions involving $\sqrt{(x-a)(x-b)}$?
3. What role does completing the square play in evaluating integrals?
4. How does the $\arctan$ term appear in integrals involving quadratic denominators?
5. Would you like a graphical explanation of how substitutions transform the integral?

Tip: Always try completing the square for quadratic expressions to simplify integration involving trigonometric or logarithmic terms.