We are given the integral:

$I = \int 12x^3 (6x^2 + 1)^{12} \, dx$

Step 1: Choosing $u$ and $dv$

To apply integration by parts, we use the formula:

$\int u \, dv = uv - \int v \, du$

We split the given function into parts:

• Let $u = 12x^3$, so that $du = 36x^2 \, dx$.
• Let $dv = (6x^2 + 1)^{12} dx$, which we will integrate.

Step 2: Finding $v$

To integrate $dv = (6x^2 + 1)^{12} dx$, we use substitution.

Let: $t = 6x^2 + 1$ Then: $dt = 12x \, dx$ Rewriting: $dx = \frac{dt}{12x}$

Since we need to integrate $(6x^2 + 1)^{12}$, we substitute:

$v = \int (6x^2 + 1)^{12} dx$

Using $t = 6x^2 + 1$, we rewrite the integral as:

$\int t^{12} \cdot \frac{dt}{12x}$

We still have $x$ in the denominator, which complicates things. Instead of integration by parts, a better approach is substitution.

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Alternative Approach: Direct Substitution

Let $t = 6x^2 + 1$, so $dt = 12x \, dx$.

Rewriting the integral:

$I = \int 12x^3 (6x^2 + 1)^{12} dx$

Rewriting $x^3$ in terms of $t$:

$x^3 = x \cdot x^2 = x \cdot \frac{t - 1}{6}$

Substituting $dx = \frac{dt}{12x}$:

$I = \int 12x \cdot \frac{t - 1}{6} \cdot t^{12} \frac{dt}{12x}$

Simplifying:

$I = \int \frac{12x (t - 1) t^{12}}{6} \cdot \frac{dt}{12x}$

$I = \int \frac{(t - 1) t^{12}}{6} \cdot \frac{dt}{1}$

$I = \int \frac{(t - 1) t^{12}}{6} dt$

Distribute:

$I = \frac{1}{6} \int (t^{13} - t^{12}) dt$

Now, integrate:

$I = \frac{1}{6} \left[ \frac{t^{14}}{14} - \frac{t^{13}}{13} \right] + C$

$I = \frac{1}{6} \left( \frac{(6x^2+1)^{14}}{14} - \frac{(6x^2+1)^{13}}{13} \right) + C$

$I = \frac{1}{84} (6x^2+1)^{14} - \frac{1}{78} (6x^2+1)^{13} + C$

Final Answer:

$\frac{1}{84} (6x^2+1)^{14} - \frac{1}{78} (6x^2+1)^{13} + C$

Would you like a step-by-step breakdown of substitution?

Related Questions:

1. How do you determine when to use integration by parts vs. substitution?
2. What is a general rule for integrating polynomials inside a power function?
3. How does substitution simplify integration of composite functions?
4. Can we solve this integral using reduction formulas?
5. How does Leibniz's rule apply to differentiation under the integral sign?

Tip:

If a function is a product of a polynomial and a power of a composite function, substitution is often more effective than integration by parts.