# Given f'(x)=(3x+1)^(-1/5), what is f(x)?

*print*Print*list*Cite

### 2 Answers

We have f'(x) = (3x + 1)^(-1/5). To find f(x) we need to find the integral of f'(x)

Int[ f'(x) dx]

=> Int [ (3x + 1)^(-1/5) dx]

let 3x + 1 = y

dy/3 = dx

=> Int [(1/3)* y^(-1/5) dy]

=> [(1/3)*y^(-1/5 + 1)] / (-1/5 + 1)

=> [(1/3)y^(4/5)]/(4/5)

=> (5/12)*y^(4/5)

substitute y = 3x + 1

=> (5/12)*(3x + 1)^(4/5) + C

**The function f(x) = (5/12)*(3x + 1)^(4/5) + C**

We'll take the inverse operation and we'll integrate the given function:

Int (3x+1)^(-1/5) dx = f(x)

We'll replace the expression 3x + 1 by the variable t.

3x + 1 = t

We'll differentiate both sides:

3dx = dt

dx = dt/3

We'll re-write the integral having as variable t:

Int (3x+1)^(-1/5) dx = Int t^(-1/5)*(dt/3)

We'll evaluate the integral

Int t^(-1/5)*(dt/3) = (1/3)*t^(-1/5 + 1)/(-1/5 + 1) + C

Int t^(-1/5)*(dt/3) = (1/3)*t^(4/5)/(4/5) + C

Int t^(-1/5)*(dt/3) = 5*t^(4/5)/12 + C

**The requested function is f(x) = 5*(3x+1)^(4/5)/12 + C**