Exercise 7.4 - Question 6
Problem
$$\int \frac{x^2}{1 - x^6} \, dx$$
Step-by-Step Solution
Step 1: Recognize the Pattern
Notice that:
$$1 - x^6 = 1 - (x^3)^2$$
And the derivative of x³ is 3x², which relates to our numerator x².
Step 2: Set Up Substitution
Let:
$$u = x^3$$
Then:
$$du = 3x^2 \, dx \quad \Rightarrow \quad x^2 \, dx = \frac{du}{3}$$
Step 3: Substitute
$$\int \frac{x^2}{1 - x^6} \, dx = \int \frac{1}{1 - u^2} \cdot \frac{du}{3}
= \frac{1}{3}\int \frac{1}{1 - u^2} \, du$$
Step 4: Apply Standard Formula
The standard formula for this form is:
$$\int \frac{1}{1 - u^2} \, du = \frac{1}{2}\ln\left|\frac{1 + u}{1 -
u}\right| + C$$
Therefore:
$$\frac{1}{3}\int \frac{1}{1 - u^2} \, du = \frac{1}{3} \cdot
\frac{1}{2}\ln\left|\frac{1 + u}{1 - u}\right| + C$$
$$= \frac{1}{6}\ln\left|\frac{1 + u}{1 - u}\right| + C$$
Step 5: Substitute Back
Replace u with x³:
$$= \frac{1}{6}\ln\left|\frac{1 + x^3}{1 - x^3}\right| + C$$
✅ Final Answer
$$\int \frac{x^2}{1 - x^6} \, dx = \frac{1}{6}\ln\left|\frac{1 + x^3}{1 -
x^3}\right| + C$$