Step 1: Interpret the function. The problem asks for the antiderivative of the function presented as two separate fractions: $\frac{3x^2+2x}{7x^2}$ and $\frac{4}{9x+2}$. Given the visual separation and the absence of an explicit operator (like $\times$ or $\cdot$), we will assume the function is the sum of these two terms. So, we need to find the integral of $f(x) = \frac{3x^2+2x}{7x^2} + \frac{4}{9x+2}$. Step 2: Simplify the first term of the function. $$ \frac{3x^2+2x}{7x^2} = \frac{x(3x+2)}{7x^2} $$ For $x \neq 0$, we can cancel an $x$ from the numerator and denominator: $$ = \frac{3x+2}{7x} $$ Now, split this into two separate fractions: $$ = \frac{3x}{7x} + \frac{2}{7x} $$ $$ = \frac{3}{7} + \frac{2}{7} \cdot \frac{1}{x} $$ Step 3: Rewrite the entire function for integration. $$ f(x) = \frac{3}{7} + \frac{2}{7x} + \frac{4}{9x+2} $$ Step 4: Integrate each term separately. We need to find $\int f(x) \,dx = \int \left( \frac{3}{7} + \frac{2}{7x} + \frac{4}{9x+2} \right) \,dx$. Integrate the first term: $$ \int \frac{3}{7} \,dx = \frac{3}{7}x $$ Integrate the second term: $$ \int \frac{2}{7x} \,dx = \frac{2}{7} \int \frac{1}{x} \,dx = \frac{2}{7} \ln|x| $$ Integrate the third term: For $\int \frac{4}{9x+2} \,dx$, use a u-substitution. Let $u = 9x+2$. Then, $du = 9 \,dx$, which means $dx = \frac{1}{9} \,du$. Substitute these into the integral: $$ \int \frac{4}{u} \left(\frac{1}{9} \,du\right) = \frac{4}{9} \int \frac{1}{u} \,du $$ $$ = \frac{4}{9} \ln|u| $$ Substitute back $u = 9x+2$: $$ = \frac{4}{9} \ln|9x+2| $$ Step 5: Combine all the integrated terms and add the constant of integration, $C$. $$ \int f(x) \,dx = \frac{3}{7}x + \frac{2}{7} \ln|x| + \frac{4}{9} \ln|9x+2| + C $$ The final answer is $\boxed{\frac{3}{7}x + \frac{2}{7} \ln|x| + \frac{4}{9} \ln|9x+2| + C}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.