Question

evaluate the integral \\(\int x^3 (x^4 - 9)^6 dx\\) by making the substitution \\(u = x^4 - 9\\). \\(\square + c\\) note: your answer should be in terms of \\(x\\) and not \\(u\\).

Explanation:

Step1: Define substitution and find $du$

Let $u = x^4 - 9$. Differentiate with respect to $x$:
$$\frac{du}{dx} = 4x^3 \implies du = 4x^3 dx \implies x^3 dx = \frac{1}{4}du$$

Step2: Rewrite integral in terms of $u$

Substitute into the original integral:
$$\int x^3 (x^4 - 9)^6 dx = \int u^6 \cdot \frac{1}{4}du = \frac{1}{4}\int u^6 du$$

Step3: Integrate with respect to $u$

Use the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + C$ for $n
eq -1$:
$$\frac{1}{4} \cdot \frac{u^{7}}{7} + C = \frac{u^7}{28} + C$$

Step4: Substitute back to $x$

Replace $u$ with $x^4 - 9$:
$$\frac{(x^4 - 9)^7}{28} + C$$

Answer:

$\frac{(x^4 - 9)^7}{28}$