Question: Find the following indefinite integral:
$$\int 12x\sqrt{2x^2-32} \, dx$$
Options:
1. $$2(2x^2-32)^{3/2} + c$$
2. $$(4x^2-64)^{3/2} + c$$
3. $$6x^2(2x^2-32)^{3/2} + c$$
4. $$\frac{9}{2}(2x^2-32)^{3/2} + c$$
1. **State the problem:** We want to find the indefinite integral $$\int 12x\sqrt{2x^2-32} \, dx$$.
2. **Recall the formula and substitution:** Let $$u = 2x^2 - 32$$. Then, $$\frac{du}{dx} = 4x$$, so $$du = 4x \, dx$$.
3. **Rewrite the integral in terms of $$u$$:**
We have $$12x \, dx = 3 \times 4x \, dx = 3 \, du$$.
So the integral becomes:
$$\int 12x \sqrt{2x^2 - 32} \, dx = \int 3 \sqrt{u} \, du = 3 \int u^{1/2} \, du$$.
4. **Integrate:**
$$3 \int u^{1/2} \, du = 3 \times \frac{u^{3/2}}{\frac{3}{2}} = 3 \times \frac{2}{3} u^{3/2} = 2 u^{3/2} + c$$.
5. **Substitute back $$u$$:**
$$2 (2x^2 - 32)^{3/2} + c$$.
6. **Conclusion:** The correct answer is option 1: $$2(2x^2-32)^{3/2} + c$$.