Logarithms as anti-derivatives. $$ \int \frac{-6}{x(\ln x)^{2}} d x=\square $$ Hint: Use the natural log function and substitution.

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To evaluate the integral:

$$
\int \frac{-6}{x(\ln x)^{2}} \, dx
$$

**Step 1: Substitution**

Let $$ u = \ln x $$. Then, the derivative of $$ u $$ with respect to $$ x $$ is:

$$
\frac{du}{dx} = \frac{1}{x} \quad \Rightarrow \quad du = \frac{1}{x} \, dx \quad \Rightarrow \quad dx = x \, du
$$

**Step 2: Rewrite the Integral in Terms of $$ u $$**

Substituting $$ u = \ln x $$ and $$ dx = x \, du $$ into the integral:

$$
\int \frac{-6}{x u^{2}} \cdot x \, du = \int \frac{-6}{u^{2}} \, du
$$

**Step 3: Integrate**

$$
\int \frac{-6}{u^{2}} \, du = -6 \int u^{-2} \, du = -6 \left( -u^{-1} \right) + C = \frac{6}{u} + C
$$

**Step 4: Substitute Back $$ u = \ln x $$**

$$
\frac{6}{u} + C = \frac{6}{\ln x} + C
$$

**Final Answer:**

$$
\int \frac{-6}{x(\ln x)^{2}} \, dx = \frac{6}{\ln x} + C
$$

where $$ C $$ is the constant of integration.

**Answer:** After simplifying, an antiderivative is six divided by log x, plus C. Thus,

6 ⁄ ln x + C
An antiderivative of the function $$ \frac{-6}{x(\ln x)^{2}} $$ is $$ \frac{6}{\ln x} + C $$, where $$ C $$ is the constant of integration.

Logarithms as anti-derivatives.

Hint: Use the natural log function and substitution.

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Logarithms as anti-derivatives. Hint: Use the natural log function and substitution.