Natural Logrithimic Integration problem

superevilcube

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Feb 8, 2007
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I need to integrate the following problem, and I get the right answer except there is a -27 which is not supposed to be there.

Integrate: (x½) / (x½ - 3) dx

I used u-substitution;

u = x½ - 3 ................ x½ = u+3
2(x½)du = dx
2(u+3)du = dx

I then substituted it in;

integrate: [(u+3) / (u)]*(2u+3)du

I simplified it further to this;

integrate: [2u + 12 + (18/u)] du

I finally integrated and got;

u² + 12u + 18ln|u| +c

I then substitued back for u, simplified and I got this:

x + 6(x½) - 27 + 18ln|(x½)-3| + C

This is the answer in the book except for the -27... can anyone lend me a hand?

EDIT: Wait... does the -27 + C = C? If so I'll feel like an idiot...
 
Let \(\displaystyle u=x^{\frac{1}{2}}, \;\ x=u^{2}, \;\ 2udu=dx\)

This gives:

\(\displaystyle \L\\\int\frac{u}{u-3}\cdot{2u}du\)

\(\displaystyle \L\\\int\left(\frac{18}{u-3}+2u+6\right)du\)

\(\displaystyle \L\\18ln(u-3)+u^{2}+6u\)

Resub:

\(\displaystyle \L\\18(\sqrt{x}-3)+x+6\sqrt{x}\)
 
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