The problem is to evaluate the integral:
$$
\int x\sqrt{2x^2 + 1} \, dx
$$
Let's go through the
solution step by step, verify it, and explain the reasoning.
---
🔹 Step 1: Use Substitution
We are given:
$$
\text{Let } u = 2x^2 + 1
$$
Now differentiate both sides with respect to $ x $:
$$
\frac{du}{dx} = 4x \quad \Rightarrow \quad du = 4x\,dx
$$
We want to express $ x\,dx $ in terms of $ du $. From $ du = 4x\,dx $, we can solve for $ x\,dx $:
$$
x\,dx = \frac{1}{4} du
$$
---
🔹 Step 2: Rewrite the Integral
Original integral:
$$
\int x\sqrt{2x^2 + 1} \, dx = \int \sqrt{u} \cdot x\,dx
$$
Substitute $ \sqrt{u} = (2x^2 + 1)^{1/2} $ and $ x\,dx = \frac{1}{4} du $:
$$
= \int \sqrt{u} \cdot \frac{1}{4} du = \frac{1}{4} \int u^{1/2} \, du
$$
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🔹 Step 3: Integrate
$$
\frac{1}{4} \int u^{1/2} \, du = \frac{1}{4} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{4} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{6} u^{3/2} + C
$$
---
🔹 Step 4: Substitute Back
Recall $ u = 2x^2 + 1 $, so:
$$
\frac{1}{6} u^{3/2} + C = \frac{1}{6} (2x^2 + 1)^{3/2} + C
$$
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✔ Final Answer:
$$
\boxed{\int x\sqrt{2x^2 + 1} \, dx = \frac{1}{6}(2x^2 + 1)^{3/2} + C}
$$
---
✔️ Verification
Differentiate the result to check:
$$
\frac{d}{dx} \left[ \frac{1}{6}(2x^2 + 1)^{3/2} \right] = \frac{1}{6} \cdot \frac{3}{2} (2x^2 + 1)^{1/2} \cdot (4x)
$$
$$
= \frac{1}{6} \cdot \frac{3}{2} \cdot 4x \cdot \sqrt{2x^2 + 1} = \frac{12x}{12} \cdot \sqrt{2x^2 + 1} = x\sqrt{2x^2 + 1}
$$
✔ Matches the original integrand.
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📌 Summary
- The substitution $ u = 2x^2 + 1 $ was chosen because the derivative $ du/dx = 4x $ matches a multiple of the remaining factor $ x $ in the integrand.
- We adjusted the integral by factoring out $ \frac{1}{4} $ to match $ 4x\,dx $.
- After integrating in terms of $ u $, we substituted back to get the answer in terms of $ x $.
✔ The solution provided is
correct.
Parent Tip: Review the logic above to help your child master the concept of calculus answers.
