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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