Exercise 7.4 - Question 2
Problem
$$\int \frac{1}{\sqrt{1 + 4x^2}} \, dx$$
Step-by-Step Solution
Step 1: Identify the Form
The integrand has the form:
$$\frac{1}{\sqrt{a^2 + u^2}}$$
We need to rewrite 1 + 4x² in this standard form. Notice that:
$$1 + 4x^2 = 1 + (2x)^2$$
This suggests using u = 2x.
Step 2: Set Up the Substitution
Let:
$$u = 2x$$
Then:
$$du = 2 \, dx \quad \Rightarrow \quad dx = \frac{du}{2}$$
Step 3: Substitute
Replace 2x with u and dx with du/2:
$$\int \frac{1}{\sqrt{1 + 4x^2}} \, dx = \int \frac{1}{\sqrt{1 + u^2}} \cdot
\frac{du}{2}$$
Factor out the constant:
$$= \frac{1}{2} \int \frac{1}{\sqrt{1 + u^2}} \, du$$
Step 4: Apply the Standard Formula
The standard integral formula is:
$$\int \frac{1}{\sqrt{a^2 + u^2}} \, du = \sinh^{-1}\left(\frac{u}{a}\right)
+ C = \ln\left|u + \sqrt{u^2 + a^2}\right| + C$$
With a = 1:
$$\int \frac{1}{\sqrt{1 + u^2}} \, du = \sinh^{-1}(u) + C = \ln\left|u +
\sqrt{u^2 + 1}\right| + C$$
Step 5: Apply the Formula
$$\frac{1}{2} \int \frac{1}{\sqrt{1 + u^2}} \, du = \frac{1}{2}
\sinh^{-1}(u) + C$$
Or equivalently:
$$= \frac{1}{2} \ln\left|u + \sqrt{u^2 + 1}\right| + C$$
Step 6: Substitute Back
Replace u with 2x:
$$= \frac{1}{2} \sinh^{-1}(2x) + C$$
Or in logarithmic form:
$$= \frac{1}{2} \ln\left|2x + \sqrt{4x^2 + 1}\right| + C$$
✅ Final Answer
$$\int \frac{1}{\sqrt{1 + 4x^2}} \, dx = \frac{1}{2}\sinh^{-1}(2x) + C$$
Or equivalently:
$$= \frac{1}{2}\ln\left|2x + \sqrt{1 + 4x^2}\right| + C$$
Key Concepts Used
- Standard Form: ∫1/√(a² + u²) du = sinh⁻¹(u/a) + C
- U-Substitution: To convert 4x² to u²
- Equivalent Forms: sinh⁻¹(x) = ln|x + √(x² + 1)|