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

1 Answer
Lovecraft
Jan 8, 2016

#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#

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