To solve the integral 

\[
\int \frac{1}{x^2 \sqrt{x^2 + 4}} \, dx,
\]

let’s approach this step by step.

---

### Step 1: Simplify and consider substitution

The presence of \( x^2 + 4 \) in the square root suggests a trigonometric substitution. Use \( x = 2 \tan\theta \), so:

\[
x^2 + 4 = 4 \sec^2\theta, \quad \sqrt{x^2 + 4} = 2 \sec\theta, \quad dx = 2 \sec^2\theta \, d\theta.
\]

Also, since \( x^2 = 4 \tan^2\theta \), then \( \frac{1}{x^2} = \frac{1}{4 \tan^2\theta} \).

---

### Step 2: Substituting into the integral

Substituting all expressions into the integral gives:

\[
\int \frac{1}{x^2 \sqrt{x^2 + 4}} \, dx
= \int \frac{1}{(4 \tan^2\theta)(2 \sec\theta)} (2 \sec^2\theta) \, d\theta.
\]

Simplify the terms:

\[
= \int \frac{\sec\theta}{8 \tan^2\theta} \, d\theta.
\]

---

### Step 3: Simplify and rewrite in terms of basic trigonometric integrals

Use the identity \( \tan^2\theta = \sec^2\theta - 1 \) to rewrite the denominator:

\[
= \frac{1}{8} \int \frac{\sec\theta}{\sec^2\theta - 1} \, d\theta.
\]

Rewrite as:

\[
= \frac{1}{8} \int \frac{1}{\tan^2\theta} \, d\theta.
\]

Since \( \frac{1}{\tan^2\theta} = \cot^2\theta \), the integral becomes:

\[
\frac{1}{8} \int \cot^2\theta \, d\theta.
\]

---

### Step 4: Integral of \(\cot^2\theta\)

Recall that:

\[
\cot^2\theta = \csc^2\theta - 1.
\]

Thus:

\[
\int \cot^2\theta \, d\theta = \int (\csc^2\theta - 1) \, d\theta = -\cot\theta - \theta + C.
\]

So the integral becomes:

\[
\frac{1}{8} \left( -\cot\theta - \theta \right) + C.
\]

---

### Step 5: Back-substitute \( \theta \) in terms of \( x \)

From the substitution \( x = 2 \tan\theta \), we know:

\[
\theta = \tan^{-1}\left(\frac{x}{2}\right), \quad \cot\theta = \frac{1}{\tan\theta} = \frac{2}{x}.
\]

Substitute these back:

\[
-\cot\theta = -\frac{2}{x}, \quad -\theta = -\tan^{-1}\left(\frac{x}{2}\right).
\]

Thus, the integral is:

\[
\int \frac{1}{x^2 \sqrt{x^2 + 4}} \, dx = \frac{1}{8} \left( -\frac{2}{x} - \tan^{-1}\left(\frac{x}{2}\right) \right) + C.
\]

Simplify:

\[
\boxed{\int \frac{1}{x^2 \sqrt{x^2 + 4}} \, dx = -\frac{1}{4x} - \frac{1}{8} \tan^{-1}\left(\frac{x}{2}\right) + C.}
\]

---

Would you like me to clarify any step or expand on the substitution? 

### Related Questions:
1. What are common trigonometric substitutions for similar integrals?
2. How can \(\tan^{-1}(x)\) derivatives and integrals help in practical problems?
3. Could this integral be solved using hyperbolic substitution instead of trigonometric?
4. How does substitution simplify complex integrals involving square roots?
5. Are there other methods, like numerical integration, to evaluate this integral?

**Tip:** Always check if trigonometric substitution or partial fractions fit better for integrals with radicals.

Learn how to solve the integral ∫1 / (x^2 √(x^2 + 4)) dx using trigonometric substitution. This detailed step-by-step solution covers trigonometric identities, substitution techniques, and integration of trigonometric functions. Suitable for undergraduate calculus students.

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