Exercise 7.4 - Question 5
Problem
$$\int \frac{3x}{1 + 2x^4} \, dx$$
Step-by-Step Solution
Step 1: Analyze the Structure
Rewrite the denominator:
$$1 + 2x^4 = 1 + (\sqrt{2}x^2)^2$$
This suggests substituting u = √2 x²
Step 2: Set Up Substitution
Let:
$$u = \sqrt{2}x^2$$
Then:
$$du = 2\sqrt{2}x \, dx \quad \Rightarrow \quad x \, dx =
\frac{du}{2\sqrt{2}}$$
Step 3: Substitute
Note that 2x⁴ = u², so 1 + 2x⁴ = 1 + u²
$$\int \frac{3x}{1 + 2x^4} \, dx = 3\int \frac{x \, dx}{1 + u^2}$$
$$= 3\int \frac{1}{1 + u^2} \cdot \frac{du}{2\sqrt{2}} =
\frac{3}{2\sqrt{2}}\int \frac{1}{1 + u^2} \, du$$
Step 4: Apply Standard Formula
$$\int \frac{1}{1 + u^2} \, du = \tan^{-1}(u) + C$$
So:
$$\frac{3}{2\sqrt{2}}\int \frac{1}{1 + u^2} \, du =
\frac{3}{2\sqrt{2}}\tan^{-1}(u) + C$$
Step 5: Substitute Back
Replace u with √2 x²:
$$= \frac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2}x^2) + C$$
✅ Final Answer
$$\int \frac{3x}{1 + 2x^4} \, dx = \frac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2}x^2)
+ C$$