So in this problem, we can take each piece of it and find each respective antiderivative. We start with #int1/(9x+2)#. We use u-substitution and set our #u=9x+23# and our #du=9dx# which can be rewritten as #1/9du=dx# so when we substitute we can pull out the constant and integrate #(1/9)int1/udu# which becomes #(1/9)lnabs(u)# and then we plug in our u to get the answer of #(1/9)lnabs(9x+2)# for the first part.
Next up we have #int30x^5dx#. 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 #30/6# to get 5 and our final answer for this is #5x^6#.
Lastly, we take the integral of #intsin(4x)#. For this we will once again turn to u-substitution to get our answer. #u=4x# and #du=4dx# which is rewritten as #(1/4)du=dx# so we can pull the constant (1/4) out front and integrate #intsin(u)du# which is #-cos(u)# and we plug in u and multiply by the constant we pulled out front to give us #(-1/4)cos(4x)# and then we put all the parts together and add the integrating constant #C# at the end to give us the answer #f(x)=1/9ln(9x+2)+5x^6-1/4cos(4x)+C#.
