Question #577a3

1 Answer
Oct 13, 2017

f(x)=1/9ln(9x+2)+5x^6-1/4cos(4x)+C

Explanation:

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.