Mathematics OpenStudy (anonymous):

Integrate (2x+5)/(x^2+2x+2)dx OpenStudy (anonymous):

partial fractions or u-sub? OpenStudy (saifoo.khan):

$\Huge \frac{\cancel{new-trick}}{\cancel{new-trick}}$ OpenStudy (anonymous):

how did i guess? OpenStudy (anonymous):

i tried partial fractions but the denomintor doesnt factorise OpenStudy (anonymous):

Let's try substitution $$\alpha(\frac{d}{dx}(x^2 +2x+2)) + \beta = 2x+5$$ OpenStudy (anonymous):

damn. well it does if you use complex numbers and then contour, but you are right, so i guess it is a u - sub. OpenStudy (anonymous):

so u=x^2+2x+2? or do i need to make the fraction into a different form? OpenStudy (anonymous):

$\int \frac{2x+3}{x^2+2x+2}dx+\int\frac{2}{x^2+2x+2}dx$ OpenStudy (anonymous):

i think that is what ishaan had yes? OpenStudy (anonymous):

right, i'll work with that. thank you OpenStudy (anonymous):

wait hold on OpenStudy (anonymous):

maybe it should be 2x+2 instead of 2x+3? OpenStudy (anonymous):

should be $\int \frac{2x+2}{x^2+2x+2}dx+\int\frac{3}{x^2+2x+2}dx$ OpenStudy (anonymous):

yeah i screwed up it is early OpenStudy (anonymous):

so do you get ln(x^2+2x+2)+3ln(x^2+2x+2)+c OpenStudy (anonymous):

Let me check OpenStudy (anonymous):

actualy thats wrong sorry OpenStudy (anonymous):

ln(x^2+2x+2)=-3/x+6ln(x)+6x+c OpenStudy (anonymous):

ah $$log(2x^2 +2x +2) + 3 tan^{-1}(x+1) +C$$ is what I get OpenStudy (anonymous):

how did u get the tan bit :( OpenStudy (anonymous):

$x^2 + 2x + 1 + 1 = (x +1)^2 + 1$ OpenStudy (anonymous):

The second part is $$\int \frac{3}{(x+1)^2 +1}dx$$ OpenStudy (anonymous):

is there no way i could just take the quadratic up and get $\int\limits_{?}^{?}3x ^{-2}+6x ^{-1}+6 dx$ OpenStudy (anonymous):

oh no because the algebra is wrong OpenStudy (anonymous):

$\frac{3}{100+20+1}\neq \frac{3}{100}+\frac{3}{20}+\frac{3}{1}$ OpenStudy (anonymous):

oh yeah lol now i feel dumb OpenStudy (amistre64):

if you take the quad up; you could try IBPs OpenStudy (anonymous):

ishaan as the right trick. any other way will be much much more complicated OpenStudy (anonymous):

IBPs ? OpenStudy (anonymous):

IBP use ILATE lol OpenStudy (anonymous):

@priyab ishaan really has it. that is the best and snappiest way OpenStudy (anonymous):

lol I didn't knew about that. thanks OpenStudy (anonymous):

thanks guys appreciate it

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