What is the antiderivative of #ln(x)/sqrtx #?

1 Answer
sente
Dec 26, 2015

#2sqrt(x)(ln(x)-2) + C#

Explanation:

For the given function, finding the antiderivative is equivalent to finding the indefinite integral. We will proceed by applying Integration by Parts.

Let #u = ln(x)# and #dv = 1/sqrt(x)dx#

Then #du = 1/xdx# and #v = 2sqrt(x)#

From the integration by parts formula #intudv = uv - intvdu#

#intln(x)/sqrt(x)dx = 2sqrt(x)ln(x) - int2sqrt(x)*1/xdx#

#= 2sqrt(x)ln(x) - 2int1/sqrt(x)dx#

#= 2sqrt(x)ln(x) - 2(2sqrt(x)) + C#

#= 2sqrt(x)(ln(x)-2) + C#

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