How do you integrate $\int \frac{ln x}{\sqrt{x}}$ $int lnx/sqrtx$ by integration by parts method?

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Answer 1 · 2016-07-23T22:03:25

$= 2 \sqrt{x} (ln (x) - 2) + C$ $=2sqrt(x)(ln(x) - 2) + C$

We derive the IBP formula from the product rule:

Take functions $u (x), v (x)$ $u(x), v(x)$

$(uv)' = u'v + uv'$

$implies uv' = (uv)' - u'v$

$therefore int uv' dx = uv - int u'v dx$

In this case, we have to take $u(x) = ln(x) and v(x) = x^(-1/2)$ because we cannot integrate the natural log function directly.

Obtain:

$u'(x) = 1/x and v(x) = 2x^(1/2)$

So $int ln(x)/(sqrt(x))dx = 2x^(1/2)ln(x) - 2int x^(1/2)*1/x dx$

$= 2x^(1/2)ln(x) - 2 int x^(-1/2) dx$

$= 2x^(1/2)ln(x) - 4x^(1/2) + C$

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

How do you integrate int lnx/sqrtx∫lnx√xint lnx/sqrtx by integration by parts method?