How do you integrate $int sqrtxlnx$ by integration by parts method?

Question

1673 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-29T02:22:16

$intsqrt(x)ln(x)dx=2/3x^(3/2)(ln(x)-2/3)+C$

Let $u = ln(x) => du = 1/xdx$
and $dv = x^(1/2)dx => v = 2/3x^(3/2)$

Using the integration by parts formula $intudv = uv-intvdu$

$intsqrt(x)ln(x)dx = intln(x)x^(1/2)dx$

$=2/3x^(3/2)ln(x)-int2/3x^(3/2)*1/xdx$

$=2/3x^(3/2)ln(x)-2/3intx^(1/2)dx$

$=2/3x^(3/2)ln(x)-2/3(2/3x^(3/2))+C$

$=2/3x^(3/2)(ln(x)-2/3)+C$

How do you integrate int sqrtxlnxint sqrtxlnx by integration by parts method?