How do you find the integral $ln x / x^(1/2)$ ?

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Answer 1 · 2018-04-10T13:23:13

$intlnx/x^(1/2)dx=2sqrtxlnx-4sqrtx+C$

Rewrite the integrand, and integrate by parts:

$intlnx/x^(1/2)dx=intx^(-1/2)lnxdx$

$u=lnx$

$du=x^-1dx$

$dv=x^(-1/2)dx$

$v=intx^(-1/2)dx=2x^(1/2)$

$uv-intvdu=2sqrtxlnx-2intx^(1/2)x^-1dx$

$=2sqrtxlnx-2intx^(-1/2)dx=2sqrtxlnx-4sqrtx+C$

So,

$intlnx/x^(1/2)dx=2sqrtxlnx-4sqrtx+C$

Answer 2 · 2018-04-10T13:26:03

$2sqrtxInx-4sqrtx + C$

Let $u=Inx$
and $dv=x^(-1/2)$ SO $v=2sqrtx$

$int(Inx)/x^(1/2) dx$ = $2sqrtxtimes Inx - int 2sqrtx times 1/x$

= $2sqrtxInx-2intx^(-1/2) dx$

= $2sqrtxInx-2times2sqrtx + C$

= $2sqrtxInx-4sqrtx + C$

How do you find the integral ln x / x^(1/2)ln x / x^(1/2)?