How do you find the integral of $\sqrt{t} \cdot ln (t) d t$ $sqrt (t)*ln (t) dt$ ?

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How do you find the integral of $\sqrt{t} \cdot ln (t) d t$ $sqrt (t)*ln (t) dt$ ?

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Answer 1 · 2015-09-04T11:08:53

use integration by parts to find $\int \sqrt{t} ln (t) d t = \frac{2}{3} t^{\frac{3}{2}} (ln (t) - \frac{2}{3}) + C$ $intsqrt(t)ln(t)dt=2/3t^(3/2)(ln(t)-2/3)+C$

Explanation:

We use integration by parts by noting that $\sqrt{t} = \frac{d}{d t} (\frac{2}{3} t^{\frac{3}{2}})$ $sqrt(t)=d/dt(2/3t^(3/2))$ .
Therefore
$\int \sqrt{t} ln (t) d t = \frac{2}{3} \int ln (t) d (t^{\frac{3}{2}}) = \frac{2}{3} (ln (t) t^{\frac{3}{2}} - \int t^{\frac{3}{2}} d ln (t))$ $intsqrt(t)ln(t)dt=2/3intln(t)d(t^(3/2))=2/3(ln(t)t^(3/2)-intt^(3/2)dln(t))$ .
Since $\frac{d}{d t} (ln (t)) = \frac{1}{t}$ $d/dt(ln(t))=1/t$ this gives
$\int \sqrt{t} ln (t) d t = \frac{2}{3} (ln (t) t^{\frac{3}{2}} - \int \frac{t^{\frac{3}{2}}}{t} d t) = \frac{2}{3} (ln (t) t^{\frac{3}{2}} - \int t^{\frac{1}{2}} d t) = \frac{2}{3} (ln (t) t^{\frac{3}{2}} - \frac{2}{3} t^{\frac{3}{2}}) + C = \frac{2}{3} t^{\frac{3}{2}} (ln (t) - \frac{2}{3}) + C$ $intsqrt(t)ln(t)dt=2/3(ln(t)t^(3/2)-intt^(3/2)/tdt)=2/3(ln(t)t^(3/2)-intt^(1/2)dt)=2/3(ln(t)t^(3/2)-2/3t^(3/2))+C=2/3t^(3/2)(ln(t)-2/3)+C$
With C the integration constant.

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How do you find the integral of $\sqrt{t} \cdot ln (t) d t$ $sqrt (t)*ln (t) dt$ ?

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