# Integral

• Feb 18th 2010, 10:43 AM
naomi
Integral
Hi,
I am trying to solve ∫ln(t)^2 dt. I already solved part of the integral, however I don't know how to complete the exercise. Can someone help me? Thanks.

∫ln(t)^2 dt
u=ln(t)^2 dv=dt
du=2ln(t)*(1/t)dt v =t

t*ln(t)^2-∫t*2ln(t)*(1/t)dt
t*ln(t)^2-∫(t*2ln(t)/t) dt
t*ln(t)^2-∫(2ln(t)) dt
• Feb 18th 2010, 10:57 AM
icemanfan
Using the fact that $\displaystyle \frac{d}{dt} t \ln t = 1 + \ln t$,

we have:

$\displaystyle \int 2 \ln t \cdot dt = 2 \int \ln t \cdot dt = 2 \left( \int (1 + \ln t) dt - \int dt \right) = 2t \ln t - 2 \int dt$

Does this help?
• Feb 18th 2010, 11:02 AM
chisigma
Integrating by parts You obtain...

$\displaystyle \int \ln^{2} t \cdot dt = t\cdot \ln^{2} t - 2\cdot \int \ln t\cdot dt$ (1)

... and integrating by parts again You obtain...

$\displaystyle \int \ln t\cdot dt = t\cdot \ln t - \int dt= t\cdot \ln t - t + c$ (2)

... so that is...

$\displaystyle \int \ln^{2} t \cdot dt = t\cdot \ln^{2} t -2\cdot t\cdot \ln t +2\cdot t + c$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Feb 19th 2010, 10:52 AM
naomi
Thanks