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I am trying to solve the following integral:

$$\int ln[(x+2)^{x+5}] dx$$

I'm not entirely sure how to go about this. Since I know that $\int udv = uv - \int vdu$ I started by assigning $u = ln[(x+2)^{x+5}]$ and $dv = dx$. I'm now left with $xln[(x+2)^{x+5}] - \int\frac{x^2+5x}{x+2}+xln(x+2)dx$.

I attempted to continue the integration, but I came up with some nonsensical gibberish.

What I don't know is if I'm taking the wrong approach entirely or if there is just something I'm messing up along the way.

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    $\begingroup$ What you've written seems correct. Now solve $\int \frac{x^2 + 5x}{x+2}\mathrm{d}x = \int \left( x + 3 - \frac{ 6}{x+2}\right) \mathrm{d}x$, and solve $\int x \log (x+2) \mathrm{d}x$ by IBP again, with $\mathrm{d}v = x\mathrm{d}x$. Note that all this would ahve been much easier to write if you'd started with the fact that $\log a^b = b \log a$. $\endgroup$ – stochasticboy321 May 4 '16 at 1:28
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Did you try solving it this way $$ \begin{align} \int ln[(x+2)^{x+5}] dx &= \int (x+5)ln(x+2) dx \\ &= (x^2/2 + 5x)ln(x+2) - \int (x^2/2 + 5x)/(x+2) dx \\ \end{align} $$ Call the integral on the right J $$ \begin{align} J & = \int (x^2/2 + 5x)/(x+2) dx \\ &= 1/2\int (((x^2 + 4x + 4) + (6x + 12) - 4 - 12)/(x+2)) dx \\ &= 1/2 \int (x + 2)dx + 3x - 8 ln(x+2) \\ &= 1/2 (x^2/2 + 2x) + 3x - 8 ln(x+2) \end{align} $$

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