# Integrate:∫(t-1)ln(t)dt

• Jan 14th 2011, 10:44 AM
mike789
Integrate:∫(t-1)ln(t)dt
∫(t-1)ln(t)dt

U=lnt dv/dx = t-1
Du/dx = 1/t V = ((t^2)/2)-t

=lnt((t^2/2)-t)- ∫((t^2/2)-t)*1/t

After doing the calculations I got:
((t^2/2)-t)lnt-(t^2/4)+t/2

((t^2/2)-t)lnt-(t^2/4)+t

Any help..plz?
• Jan 14th 2011, 10:56 AM
Pranas
Quote:

Originally Posted by mike789
∫(t-1)ln(t)dt

U=lnt dv/dx = t-1
Du/dx = 1/t V = ((t^2)/2)-t

=lnt((t^2/2)-t)- ∫((t^2/2)-t)*1/t

After doing the calculations I got:
((t^2/2)-t)lnt-(t^2/4)+t/2

((t^2/2)-t)lnt-(t^2/4)+t

Any help..plz?

$\displaystyle \displaystyle $\int {(t - 1)\ln t \cdot {\mkern 1mu} dt}$$

Notations
$\displaystyle \displaystyle $u = \ln t \Rightarrow du = \frac{{dt}}{t}$$
$\displaystyle \displaystyle $dv = (t - 1) \cdot dt \Rightarrow v = \frac{{{t^2}}}{2} - t$$

Gives us
$\displaystyle \displaystyle $\int {(t - 1)\ln t \cdot {\mkern 1mu} dt} = u \cdot v - \int {v \cdot du = } (\frac{{{t^2}}}{2} - t) \cdot \ln t - \int {(\frac{{{t^2}}}{2} - t)} \frac{{dt}}{t} =$$
$\displaystyle \displaystyle $= (\frac{{{t^2}}}{2} - t) \cdot \ln t - \int {(\frac{t}{2} - 1)} \cdot dt = (\frac{{{t^2}}}{2} - t) \cdot \ln t - \frac{{{t^2}}}{4} + t + C$$
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