Let's go step-by-step to solve the integral:

$\int \frac{1}{x(x^2 + 5)} \, dx$

Step 1: Use substitution

We can use partial fractions to simplify the integral. Let's first decompose the fraction.

1. The expression $\frac{1}{x(x^2 + 5)}$ can be split into: $\frac{A}{x} + \frac{Bx + C}{x^2 + 5}$ Multiply both sides by $x(x^2 + 5)$ to get: $1 = A(x^2 + 5) + (Bx + C)x$ Expanding: $1 = A(x^2) + 5A + Bx^2 + Cx$ Group the terms: $1 = (A + B)x^2 + Cx + 5A$ Now, equate coefficients of powers of $x$:

   • For $x^2$: $A + B = 0$
   • For $x^1$: $C = 0$
   • Constant: $5A = 1$

   From $5A = 1$, we get: $A = \frac{1}{5}$

   From $A + B = 0$, we get: $B = -\frac{1}{5}$

   And $C = 0$.

   So, the partial fraction decomposition is: $\frac{1}{x(x^2 + 5)} = \frac{1/5}{x} - \frac{1/5}{x^2 + 5}$

Step 2: Integrate the two terms separately

Now we can integrate each term:

$\int \frac{1}{x(x^2 + 5)} \, dx = \frac{1}{5} \int \frac{1}{x} \, dx - \frac{1}{5} \int \frac{1}{x^2 + 5} \, dx$

1. The first integral is straightforward: $\frac{1}{5} \int \frac{1}{x} \, dx = \frac{1}{5} \ln |x|$

2. The second integral: $\int \frac{1}{x^2 + 5} \, dx$ This is a standard form, where $\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right)$.

   Here, $a^2 = 5$, so $a = \sqrt{5}$: $\int \frac{1}{x^2 + 5} \, dx = \frac{1}{\sqrt{5}} \tan^{-1} \left( \frac{x}{\sqrt{5}} \right)$

Step 3: Final answer

Now substitute back into the original expression:

$\int \frac{1}{x(x^2 + 5)} \, dx = \frac{1}{5} \ln |x| - \frac{1}{5 \sqrt{5}} \tan^{-1} \left( \frac{x}{\sqrt{5}} \right) + C$

Summary:

The integral is:

$\int \frac{1}{x(x^2 + 5)} \, dx = \frac{1}{5} \ln |x| - \frac{1}{5 \sqrt{5}} \tan^{-1} \left( \frac{x}{\sqrt{5}} \right) + C$

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Would you like more details on any step or have any questions?

5 Related Questions:

1. How does partial fraction decomposition work in more complex cases?
2. Can we integrate this using substitution only instead of partial fractions?
3. What is the geometric interpretation of the $\tan^{-1}$ term in integrals?
4. How do we handle integrals of higher degree polynomials in the denominator?
5. What other methods exist for solving rational function integrals?

Tip:

When dealing with tricky integrals involving rational functions, always check if partial fractions can simplify the process.