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

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Answer 1 · 2016-01-08T11:03:25

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

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 $intln(x^n)dx = n(ln(x)-x) + c$

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