To solve the integral \( \int \frac{3ax}{b^2 + c^2 x^2} \, dx \), let's proceed with a substitution to simplify the integrand. 

Consider the substitution \( u = c^2 x^2 + b^2 \). Then we have:

\[ du = 2c^2 x \, dx \]
\[ dx = \frac{du}{2c^2 x} \]

Rewriting the integrand in terms of \( u \), we have:

\[ \int \frac{3ax}{b^2 + c^2 x^2} \, dx = \int \frac{3a}{u} \cdot \frac{du}{2c^2 x} \]

Since \( u = b^2 + c^2 x^2 \), we can express \( x \) in terms of \( u \):

\[ x = \sqrt{\frac{u - b^2}{c^2}} \]

Substituting this back into the integral:

\[ \int \frac{3a}{u} \cdot \frac{du}{2c^2 \sqrt{\frac{u - b^2}{c^2}}} \]

Simplify the expression:

\[ \int \frac{3a}{u} \cdot \frac{du}{2c \sqrt{u - b^2}} \]

Recognize that \( \sqrt{u - b^2} = \sqrt{\frac{u - b^2}{c^2}} \cdot c = \sqrt{u - b^2} \cdot \frac{1}{\sqrt{c^2}} \cdot c = \sqrt{u - b^2} \):

\[ \int \frac{3a}{u} \cdot \frac{du}{2c (u - b^2)^{1/2}} \]

Integrate:

\[ \frac{3a}{2c} \int \frac{du}{u (u - b^2)^{1/2}} \]

We need to perform a partial fraction decomposition or a suitable substitution to integrate this expression. Let's consider the substitution \( u = b^2 \sec^2 \theta \):

\[ du = 2b^2 \sec^2 \theta \tan \theta \, d\theta \]
\[ \sqrt{u - b^2} = b \tan \theta \]

Substituting these into the integral:

\[ \frac{3a}{2c} \int \frac{2b^2 \sec^2 \theta \tan \theta \, d\theta}{b^2 \sec^2 \theta \cdot b \tan \theta} = \frac{3a}{2c} \int \frac{2b^2 \sec^2 \theta \tan \theta \, d\theta}{b^3 \sec^3 \theta} = \frac{3a}{2c} \int \frac{2b^2 \sec^2 \theta \tan \theta \, d\theta}{b^3 \sec^2 \theta \sec \theta} = \frac{3a}{2c} \int \frac{2 \tan \theta \, d\theta}{b \sec \theta}\]

Simplify further:

\[ = \frac{3a}{2cb} \int \frac{2 \tan \theta \, d\theta}{\sec \theta} = \frac{3a}{cb} \int \sin \theta \, d\theta \]

Integrate \( \sin \theta \):

\[ = \frac{3a}{cb} (-\cos \theta) = -\frac{3a}{cb} \cos \theta + C \]

Finally, substitute back \( \theta = \sec^{-1} \left( \frac{\sqrt{u}}{b} \right) \):

\[ -\frac{3a}{cb} \cos (\sec^{-1} \left( \frac{\sqrt{u}}{b} \right)) + C = -\frac{3a}{cb} \cdot \frac{b}{\sqrt{u}} + C = -\frac{3a}{c} \cdot \frac{1}{\sqrt{b^2 + c^2 x^2}} + C \]

So, the integral \( \int \frac{3ax}{b^2 + c^2 x^2} \, dx \) evaluates to:

\[ -\frac{3a}{c} \cdot \frac{1}{\sqrt{b^2 + c^2 x^2}} + C \]

Would you like further details or have any questions?

Here are five questions you might consider asking:
1. How do you handle integrals with different types of substitutions?
2. Can you explain more about partial fraction decomposition?
3. What are some common techniques for solving integrals?
4. How do trigonometric substitutions work in integrals?
5. Can you show an example of a more complex integral?

**Tip:** Always check if a substitution can simplify your integral. This often turns a difficult problem into an easier one.

Learn how to solve the integral ∫ (3ax)/(b^2 + c^2x^2) dx using substitution and trigonometric methods. This problem covers advanced integration techniques, including substitution and trigonometric substitution. Suitable for advanced high school or college-level students.

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