[SOLVED] Definite integral involving parts

**Archduke01** · April 10th 2010, 07:24 PM

$\int_2^{6} \sqrt {t} ln t dt$

$u = ln t$
$dv = \sqrt{t}$
$du = 1/t$
$v = 2t^{\frac{3}{2}}$
$<br /> \frac{2 ln t*t^\frac{3}{2}}{3} - \int \frac {2t^\frac{3}{2}}{3t}$
$\frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3} \int \frac {t^\frac{3}{2}}{t}$
$\frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3}* \frac {2t^\frac{3}{2}}{3}$

Could anyone point out the mistake I made? Continuing with this, my answer turns out wrong.

EDIT: Nevermind guys, I have my answer. Sorry for the waste of topic. How do I delete my topic?

**harish21** · April 10th 2010, 07:38 PM

Originally Posted by Archduke01

$\int_2^{6} \sqrt {t} ln t dt$

$u = ln t$
$dv = \sqrt{t}$
$du = 1/t$
$v = 2t^{\frac{3}{2}}$
$<br /> \frac{2 ln t*t^\frac{3}{2}}{3} - \int \frac {2t^\frac{3}{2}}{3t}$
$\frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3} \int \frac {t^\frac{3}{2}}{t}$
$\frac{2 ln t*t^\frac{3}{2}}{3} - \frac {-2}{3}* \frac {2t^\frac{3}{2}}{3}$

Could anyone point out the mistake I made? Continuing with this, my answer turns out wrong.

The fourth line of yours should be:
$v = \frac{2}{3}t^{\frac{3}{2}}$

and you should write $du = \frac{1}{t}dt$

So,

$uv - \int v du$

$= \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \int \frac{2}{3}t^{\frac{3}{2}} \times \frac{1}{t} dt$

$= \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \frac{2}{3}\int t^{\frac{3}{2}} \times \frac{1}{t} dt$

$= \frac{2}{3}t^{\frac{3}{2}}.ln(t) - \frac{2}{3}\int t^{{\frac{3}{2}}-1} dt$

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