Today

>today
>to intelligent two study
>pic related in exam
>easy as fuck
>got confused by INT(x-1)^-2
>1h30min trying the fucking problem with integ. methods and log shit
>5 min left
>I remembered the denominators method exist
>check answer online
>(x-1)^-1
>tfw there were another question left
>tfw I might get up to 6/10 because of this problem
R*E^(10^100)

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>1h30min trying the fucking problem with integ. methods and log shit
Didn't even try a u sub? Or think about moving on to the next question?

Please tell me those polynomial are irreducible

Can group the denominator up and haul our -X^2 on the left. Too intellectual to do it fully tb.h

[math]\frac{2x^2+x+3}{x^3-x^2-x+1} = \frac{2x^2+x+3}{(x+1)(x-1)^2}\\
\frac{2x^2+x+3}{(x+1)(x-1)^2} = \frac{A_0}{x+1}+\frac{A_1}{x-1}+\frac{A_2}{(x-1)^2}\\
2x^2+x+3 = A_0(x-1)^2+A_1(x-1)(x+1)+A_2(x+1)\\
2x^2+x+3 = A_0(x-1)^2+A_1(x^2-1)+A_2(x+1)\\
\left\{ \begin{array}{ccc}
x=+1 & 2+1+3 = 2A_2 \\
x=-1 & 2-1+3 = 4A_0 \end{array} \right.\\
\left\{ \begin{array}{ccc}
x=+1 & 6 = 2A_2 \\
x=-1 & 4 = 4A_0 \end{array} \right.\\
\left\{ \begin{array}{ccc}
x=+1 & 3 = A_2 \\
x=-1 & 1 = A_0 \end{array} \right.\\
x=0 \hspace{4 mm} 0+0+3 = A_0 -A_1 + A_2\\
x=0 \hspace{4 mm} 3 = 1 -A_1 + 3\\
x=0 \hspace{4 mm} -1 = -A_1\\
x=0 \hspace{4 mm} A_1 = 1\\
\frac{2x^2+x+3}{(x+1)(x-1)^2} = \frac{1}{x+1}+\frac{1}{x-1}+\frac{3}{(x-1)^2}\\
\int \frac{1}{x+1}\,dx + \int \frac{1}{x-1}\,dx + \int \frac{3}{(x-1)^2}\,dx\\
ln(x+1) + ln(x-1) + \int \frac{3}{u^2}\,du ; \hspace{4 mm} u=x+1, dx=du\\
ln(x+1) + ln(x-1) + \frac{3}{u}\frac{1}{-1} + C\\
ln(x+1) + ln(x-1) - \frac{3}{x+1} + C\\
[/math]

>cubic
>irreducible

[math]u=x-1,dx=du\\
ln(|x^2-1|)-\frac{3}{x-1}+C[/math]

>tfw to intelligent two skip questions

Because most physics is dull

Did you do that first factorization by inspection? I totally missed that