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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Sign up to join this communityHow 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.
$$ I = \int \frac{1-x}{x^2+2x+2} dx = \int \frac{1-x}{(x+1)^2+1} dx$$ Then substitute $y = x+1, dy = dx$, to get $$ I = \int\frac{2-y}{y^2+1}dy = 2 \arctan(x+1)-\frac{1}{2}\log(x^2+2x+2)+C$$
Try to consider it as $(v+1)^2+1$ and then you have $$\dfrac{1-v}{v^2+2v+2}=\dfrac{-0.5(2v+2)}{v^2+2v+2}+\dfrac{2}{\underbrace{v^2+2v+2}_{(v+1)^2+1}}$$