$$ \int\left(4-3x\right)^5 $$$$ \differentialD x $$

To solve the integral:

\[
I = \int (4 - 3x)^5 \, dx
\]

We can apply **substitution** to simplify it. Let:

\[
u = 4 - 3x
\]

Now differentiate $ u $ with respect to $ x $:

\[
\frac{du}{dx} = -3 \quad \text{or} \quad dx = \frac{du}{-3}
\]

Substitute these into the original integral:

\[
I = \int u^5 \cdot \frac{du}{-3}
\]

This simplifies to:

\[
I = -\frac{1}{3} \int u^5 \, du
\]

Now, integrate $ u^5 $:

\[
I = -\frac{1}{3} \cdot \frac{u^6}{6} + C = -\frac{u^6}{18} + C
\]

Finally, substitute back $ u = 4 - 3x $:

\[
I = -\frac{(4 - 3x)^6}{18} + C
\]

Thus, the solution to the integral is:

\[
\boxed{-\frac{(4 - 3x)^6}{18} + C}
\]

Would you like further details or have any questions?

Here are 5 related questions for you to explore:
1. How do you apply substitution in integrals with complex expressions?
2. Can substitution be used for definite integrals, and how does it change the bounds?
3. What is the general form of a power rule for integrals?
4. How do you reverse the process of differentiation (i.e., antidifferentiation)?
5. What are some common mistakes to avoid when using substitution in integrals?

**Tip:** Always adjust the differential when substituting to avoid mistakes in integration.

Solve the integral ∫(4 - 3x)^5 dx step-by-step using the substitution method. This solution applies the substitution rule and the power rule for integration. Suitable for Grades 11-12 students learning integral calculus.

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