Exercise 7.4 - Question 1
Problem
$$\int \frac{3x^2}{x^6 + 1} \, dx$$
Step-by-Step Solution
Step 1: Identify the Pattern
Look at the integrand carefully:
$$\frac{3x^2}{x^6 + 1}$$
Notice that:
- The denominator is x⁶ + 1, which can be written as (x³)² + 1
- The numerator 3x² is exactly the derivative of x³
This suggests a u-substitution where u = x³
Step 2: Set Up the Substitution
Let:
$$u = x^3$$
Differentiate both sides:
$$\frac{du}{dx} = 3x^2$$
Therefore:
$$du = 3x^2 \, dx$$
Step 3: Rewrite the Integral
Substitute u = x³ and du = 3x² dx into the original integral:
$$\int \frac{3x^2}{x^6 + 1} \, dx = \int \frac{1}{u^2 + 1} \, du$$
This is because:
- x⁶ + 1 = (x³)² + 1 = u² + 1
- 3x² dx = du
Step 4: Apply the Standard Formula
We now have a standard integral form:
$$\int \frac{1}{u^2 + 1} \, du$$
This is a well-known integral. The standard formula is:
$$\int \frac{1}{u^2 + 1} \, du = \tan^{-1}(u) + C$$
Step 5: Substitute Back
Replace u with x³ to get the final answer:
$$= \tan^{-1}(x^3) + C$$
✅ Final Answer
$$\int \frac{3x^2}{x^6 + 1} \, dx = \tan^{-1}(x^3) + C$$
Verification
To verify, differentiate the answer:
$$\frac{d}{dx}[\tan^{-1}(x^3)] = \frac{1}{1 + (x^3)^2} \cdot 3x^2 =
\frac{3x^2}{1 + x^6}$$
This matches our original integrand ✓
Key Concepts Used
- U-Substitution: Recognizing that 3x² is the derivative of x³
- Standard Integral: ∫1/(u²+1)du = tan⁻¹(u) + C