How do you integrate $t^{\frac{1}{2}} \cdot ln (t)$ $t^(1/2)*ln(t)$ ?

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How do you integrate $t^{\frac{1}{2}} \cdot ln (t)$ $t^(1/2)*ln(t)$ ?

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Answer 1 · 2018-03-05T08:49:58

$I = \frac{2}{9} \cdot t^{\frac{3}{2}} [3 \cdot ln t - 2] + C$ $I=2/9*t^(3/2)[3*lnt-2]+C$

Explanation:

Integration by Parts:
$\int u \cdot v d t = u \int v d t - \int {\frac{d u}{d t} \int v d t} d t$ $color(red)(intu*vdt=uintvdt-int{(du)/(dt)intvdt}dt)$
Let, $u = ln t, and, v = t^{\frac{1}{2}} \Rightarrow \frac{d u}{d t} = \frac{1}{t}, \int v d t = \frac{t^{(\frac{1}{2}) + 1}}{(\frac{1}{2}) + 1} = \frac{t (\frac{3}{2})}{\frac{3}{2}}$ $u=lnt,and,v=t^(1/2)rArr(du)/(dt)=1/t,intvdt=t^((1/2)+1)/((1/2)+1)=(t(3/2))/(3/2)$
$I=intt^(1/2)*lntdt=lnt*intt^(1/2)dt-intd/(dt)(lnt)dt*intt^(1/2)dtrArrI=lnt*t^(3/2)/(3/2)-int1/t*t^(3/2)/(3/2)*dt=2/3*t^(3/2)lnt-2/3intt^(3/2-1)dt=2/3t^(3/2)*lnt-2/3intt^(1/2)dt=2/3t^(3/2)*lnt-2/3*t^(3/2)/(3/2)+C$
$I=2/3*t^(3/2)*lnt-4/9*t^(3/2)+C$
$I=2/9*t^(3/2)[3*lnt-2]+C$

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How do you integrate $t^{\frac{1}{2}} \cdot ln (t)$ $t^(1/2)*ln(t)$ ?

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Explanation:

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