Question

What is the antiderivative of `ln(root3(x))`?

Answer 1

`I = 1/3(ln(x)x - x) + c`

Explanation:

Use the log property that we can pass exponents to the front of the log
`ln(root(3)(x)) = ln(x)/3`

`I = intln(x)/3dx`
`I = 1/3intln(x)dx`

Say `u = ln(x)` so `du = 1/x` and `dv = 1` so `v = x`

`I = 1/3(ln(x)x - intx/xdx)`
`I = 1/3(ln(x)x - intdx)`
`I = 1/3(ln(x)x - x) + c`

In general, we can say that `intln(x^n)dx = n(ln(x)-x) + c`