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**LostInCalculus** Hello, I was hoping someone could pick off where I went wrong on these two questions. As far as I can tell I have the right answers but the program I am plugging the answers into says nay. Not sure on how to do the integral symbol on this forum so just assume it is there.

1) Integrate by partial fractions $\displaystyle -2x^2-4x+8/(X+1)(X^2+4)$

Using the partial fractions I was able to change it to $\displaystyle 2/(x+1)-4/(X^2+4)$ which is so far correct according to the program (in bx+c on the top of the second fraction the c is 0)

So I am down to integrating $\displaystyle 2/(x+1)$ and $\displaystyle -4/(x^2+4)$ . Using the natural log rule the first I figure is 2ln(x+1) and the second I solved using $\displaystyle u=x^2+4$ and getting 1/2du=x. The 1/2 comming outside of the integral it leaves 1/2 the integral of du/u, which is 1/2ln(u), which then gives you -4(1/2)ln(x^2+4) as the second integral.

Mr F says: NO! First of all, it's a standard form: $\displaystyle \int \frac{a}{x^2 + a^2} = \tan^{-1}\left( \frac{x}{a} \right)$. As to why and where your substitution goes wrong, see below.

So I figured $\displaystyle 2(ln(x+1))-2ln(x^2+4)$ should be the answer, but it comes back as wrong and I am running out of attempts with no clue where I am wrong on this.

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