# Question #577a3

$f \left(x\right) = \frac{1}{9} \ln \left(9 x + 2\right) + 5 {x}^{6} - \frac{1}{4} \cos \left(4 x\right) + C$
So in this problem, we can take each piece of it and find each respective antiderivative. We start with $\int \frac{1}{9 x + 2}$. We use u-substitution and set our $u = 9 x + 23$ and our $\mathrm{du} = 9 \mathrm{dx}$ which can be rewritten as $\frac{1}{9} \mathrm{du} = \mathrm{dx}$ so when we substitute we can pull out the constant and integrate $\left(\frac{1}{9}\right) \int \frac{1}{u} \mathrm{du}$ which becomes $\left(\frac{1}{9}\right) \ln \left\mid u \right\mid$ and then we plug in our u to get the answer of $\left(\frac{1}{9}\right) \ln \left\mid 9 x + 2 \right\mid$ for the first part.
Next up we have $\int 30 {x}^{5} \mathrm{dx}$. We add one to the exponent and divide by the reciprocal of that. $5 + 1 = 6$ for the exponent and then we divide the constant $\frac{30}{6}$ to get 5 and our final answer for this is $5 {x}^{6}$.
Lastly, we take the integral of $\int \sin \left(4 x\right)$. For this we will once again turn to u-substitution to get our answer. $u = 4 x$ and $\mathrm{du} = 4 \mathrm{dx}$ which is rewritten as $\left(\frac{1}{4}\right) \mathrm{du} = \mathrm{dx}$ so we can pull the constant (1/4) out front and integrate $\int \sin \left(u\right) \mathrm{du}$ which is $- \cos \left(u\right)$ and we plug in u and multiply by the constant we pulled out front to give us $\left(- \frac{1}{4}\right) \cos \left(4 x\right)$ and then we put all the parts together and add the integrating constant $C$ at the end to give us the answer $f \left(x\right) = \frac{1}{9} \ln \left(9 x + 2\right) + 5 {x}^{6} - \frac{1}{4} \cos \left(4 x\right) + C$.