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Integration by Parts - HSN forum

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Integration by Parts


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#1 Hev

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Posted 30 January 2007 - 01:02 PM

2002. quesion 5.

Use integration by parts to evaluate the intergral between 1 and 0 [ln(1+x)]dx

the answer is 2ln2 - 1

i got ln(1+x) - [(1+x)^2]/2

please help!! sad.gif
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Posted 30 January 2007 - 06:01 PM

Use integration by parts, so:

\int^1_0 ln (1+x) dx = \{x ln(1+x) - \int \frac{x}{x+1}\}_{x=0}^{x=1}

For the integral in here, use the substitution u=x+1:


\int^1_0 ln (1+x) dx = \{x ln(1+x) - \int \frac{u-1}{u} \} _{x=0}^{x=1}

Carry out in the integration in terms of u, then substitute back in x+1 in place of u:

\int^1_0 ln (1+x) dx = \{x ln(1+x) +  ln(1+x) -x-1 \} _{x=0}^{x=1}

Put in the limits, it should come out now.

I thought this was a tricky paper. I though that q.6 was more obvious without the substition, and q.10 was quite a bit more difficult than most of the other series questions.
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