Integrate (2x+5)/(x^2+2x+2)dx
partial fractions or u-sub?
\[\Huge \frac{\cancel{new-trick}}{\cancel{new-trick}} \]
how did i guess?
i tried partial fractions but the denomintor doesnt factorise
Let's try substitution \( \alpha(\frac{d}{dx}(x^2 +2x+2)) + \beta = 2x+5\)
damn. well it does if you use complex numbers and then contour, but you are right, so i guess it is a u - sub.
so u=x^2+2x+2? or do i need to make the fraction into a different form?
\[\int \frac{2x+3}{x^2+2x+2}dx+\int\frac{2}{x^2+2x+2}dx\]
i think that is what ishaan had yes?
right, i'll work with that. thank you
wait hold on
maybe it should be 2x+2 instead of 2x+3?
should be \[\int \frac{2x+2}{x^2+2x+2}dx+\int\frac{3}{x^2+2x+2}dx\]
yeah i screwed up it is early
so do you get ln(x^2+2x+2)+3ln(x^2+2x+2)+c
Let me check
actualy thats wrong sorry
ln(x^2+2x+2)=-3/x+6ln(x)+6x+c
ah \(log(2x^2 +2x +2) + 3 tan^{-1}(x+1) +C\) is what I get
how did u get the tan bit :(
\[x^2 + 2x + 1 + 1 = (x +1)^2 + 1\]
The second part is \( \int \frac{3}{(x+1)^2 +1}dx\)
is there no way i could just take the quadratic up and get \[\int\limits_{?}^{?}3x ^{-2}+6x ^{-1}+6 dx\]
oh no because the algebra is wrong
\[\frac{3}{100+20+1}\neq \frac{3}{100}+\frac{3}{20}+\frac{3}{1}\]
oh yeah lol now i feel dumb
if you take the quad up; you could try IBPs
ishaan as the right trick. any other way will be much much more complicated
IBPs ?
IBP use ILATE lol
@priyab ishaan really has it. that is the best and snappiest way
lol I didn't knew about that. thanks
thanks guys appreciate it
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