To solve the integral $\int 9x \ln(x) \, dx$ using integration by parts, we need to select two functions $u$ and $dv$ such that:

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

We can break down the integrand $9x \ln(x)$ into a product of two functions:

• $u = \ln(x)$
• $dv = 9x \, dx$

Step 1: Differentiate and integrate

Now, we compute $du$ and $v$:

1. $u = \ln(x) \implies du = \frac{1}{x} \, dx$
2. $dv = 9x \, dx \implies v = \frac{9}{2}x^2$

Step 2: Apply the integration by parts formula

Now substitute into the formula:

$\int 9x \ln(x) \, dx = \ln(x) \cdot \frac{9}{2}x^2 - \int \frac{9}{2}x^2 \cdot \frac{1}{x} \, dx$

Simplify the second integral:

$\int 9x \ln(x) \, dx = \frac{9}{2}x^2 \ln(x) - \frac{9}{2} \int x \, dx$

Step 3: Solve the remaining integral

$\frac{9}{2} \int x \, dx = \frac{9}{2} \cdot \frac{x^2}{2} = \frac{9}{4}x^2$

Step 4: Combine terms

Now, substitute back:

$\int 9x \ln(x) \, dx = \frac{9}{2}x^2 \ln(x) - \frac{9}{4}x^2 + C$

Thus, the evaluated integral is:

$\int 9x \ln(x) \, dx = \frac{9}{2}x^2 \ln(x) - \frac{9}{4}x^2 + C$

Step 5: Check the result by differentiating

Differentiate the result to confirm:

$\frac{d}{dx} \left( \frac{9}{2}x^2 \ln(x) - \frac{9}{4}x^2 \right)$

Using the product rule on $\frac{9}{2}x^2 \ln(x)$:

$\frac{d}{dx} \left( \frac{9}{2}x^2 \ln(x) \right) = \frac{9}{2} \left( 2x \ln(x) + x \right) = 9x \ln(x) + \frac{9}{2}x$

Now differentiate $-\frac{9}{4}x^2$:

$\frac{d}{dx} \left( -\frac{9}{4}x^2 \right) = -\frac{9}{2}x$

Add the two results:

$9x \ln(x) + \frac{9}{2}x - \frac{9}{2}x = 9x \ln(x)$

Since the derivative matches the original integrand, the solution is correct.

Questions:

1. Can you explain why integration by parts works in this case?
2. What happens if we choose different functions for $u$ and $dv$?
3. How do you decide which function to assign to $u$ and $dv$ in integration by parts?
4. Can we evaluate this integral using another method, such as substitution?
5. How does the constant of integration play a role in indefinite integrals?

Tip: When applying integration by parts, choosing the correct $u$ and $dv$ can simplify the integration process significantly. Look for functions that become simpler after differentiation.