Thread: Integration of Logaritmic Functions
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Old January 2nd, 2009, 11:57 AM
Scott H Scott H is offline
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Try Integration by Substitution, which is basically the Chain Rule backward:

\int\,f'(g(x))g'(x)\,dx\,=\,f(g(x))\,+\,C

Letting u\,=\,g(x)\,=\,x^2\,+\,6x, we can see that du\,=\,g'(x)\,dx\,=\,(2x\,+\,6)\,dx\,=\,2(x\,+\,3)\,dx, so that the integral becomes

\int_{u\,=\,7}^{u\,=\,27}\,\frac{1}{2u}\,du.

Hope this helps.
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