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How does one integrate $\int \frac{1-v}{v^2+2v+2}dv$?

I tried the method of splitting the fraction into partial fractions but the denominator cannot be factorised.

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    $\begingroup$ you do not need to factor the denominator, rewrite it as $(v+1)^2+1$ $\endgroup$ – Vasya May 18 '20 at 14:12
  • $\begingroup$ @Vasya May I know what is it about the equation that makes you know rewriting as $(v+1)^2 + 1$ works? $\endgroup$ – Taenyfan May 18 '20 at 14:22
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$$ I = \int \frac{1-x}{x^2+2x+2} dx = \int \frac{1-x}{(x+1)^2+1} dx$$ Then substitute $y = x+1, dy = dx$, to get $$ I = \int\frac{2-y}{y^2+1}dy = 2 \arctan(x+1)-\frac{1}{2}\log(x^2+2x+2)+C$$

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Try to consider it as $(v+1)^2+1$ and then you have $$\dfrac{1-v}{v^2+2v+2}=\dfrac{-0.5(2v+2)}{v^2+2v+2}+\dfrac{2}{\underbrace{v^2+2v+2}_{(v+1)^2+1}}$$

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