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I don't get this integral question? The answer is supposed to be -4 + 20In2 - 6In3, but I keep getting 0
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\[2In(u-1)\sqrt{u} - 2\int\limits_{}^{}{u^{1/2}(1/(u-1))}du\] this is what I got so far
\[2\sqrt{n} In(u-1) - 2\left\{ \sqrt{n}In(u-1) - \int\limits_{}^{}In(u-1)(1/2)(u ^{-1/2})du \right\}\]
Then the 2rootIn(u-1) gets cancelled out because there are two of the same terms so I'm just left with the term 2integral(u-1)(1/2).....
Is this right?
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