To solve the given expression for the antiderivative, we break it down step by step. You are tasked with finding the antiderivative of:

$\frac{1}{\log(2)} \int \log(-0.42x + 3) \, dx$

Step 1: Integration of the Logarithmic Function

First, let's focus on the integral:

$I = \int \log(-0.42x + 3) \, dx$

We can solve this by using integration by parts. Recall the formula for integration by parts:

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

Let:

• $u = \log(-0.42x + 3)$, so $du = \frac{-0.42}{-0.42x + 3} \, dx$
• $dv = dx$, so $v = x$

Now, applying the integration by parts formula:

$I = x \log(-0.42x + 3) - \int x \cdot \frac{-0.42}{-0.42x + 3} \, dx$

Step 2: Simplification of the Remaining Integral

The remaining integral is:

$\int \frac{0.42x}{-0.42x + 3} \, dx$

We can simplify this by using a substitution:

Let $u = -0.42x + 3$, so $du = -0.42 \, dx$, and $dx = -\frac{du}{0.42}$. When we make this substitution, we get:

$\int \frac{0.42x}{-0.42x + 3} \, dx = \int \frac{3 - u}{u} \cdot \frac{-du}{0.42}$

This will break down to:

$\int \frac{3 - u}{u} \cdot \frac{-du}{0.42} = -\frac{1}{0.42} \int \left( \frac{3}{u} - 1 \right) \, du$

Now, integrating:

$= -\frac{1}{0.42} \left( 3 \ln|u| - u \right) + C$

Substitute $u = -0.42x + 3$ back into the expression:

$= -\frac{1}{0.42} \left( 3 \ln| -0.42x + 3 | - (-0.42x + 3) \right) + C$

Step 3: Putting Everything Together

Now, substituting this back into the original expression for $I$:

$I = x \log(-0.42x + 3) + \frac{1}{0.42} \left( 3 \ln| -0.42x + 3 | - (-0.42x + 3) \right) + C$

Step 4: Applying the Constant Factor $\frac{1}{\log(2)}$

Now, we multiply the whole expression by $\frac{1}{\log(2)}$ to get the final answer:

$\frac{1}{\log(2)} \left( x \log(-0.42x + 3) + \frac{1}{0.42} \left( 3 \ln| -0.42x + 3 | - (-0.42x + 3) \right) \right) + C$

Final Answer:

$\frac{1}{\log(2)} \left( x \log(-0.42x + 3) + \frac{1}{0.42} \left( 3 \ln| -0.42x + 3 | - (-0.42x + 3) \right) \right) + C$

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Let me know if you would like any further details or clarifications!

Additional Questions:

1. What is the role of integration by parts in solving logarithmic integrals?
2. How does substitution simplify integrals involving fractions and logarithms?
3. What are some common properties of logarithmic functions used in calculus?
4. How can the constant $\frac{1}{\log(2)}$ be interpreted in relation to scaling the result?
5. What techniques can be applied when dealing with more complex functions inside logarithms?

Tip:

When facing integrals involving logarithmic functions, always consider using integration by parts or substitution, as these are common strategies for simplifying such integrals.