Compute $\int\frac{1-x}{(x-2)(x+3)}$ and $\int\frac{cos(3x)}{sin(3x)}$. I have no idea how to solve these 2 integrals, I've run out of ideas. The first one especially, I can't even start, not sure how to do it. I've done this for the second one:
I introduced a new variable $t$, $3x = t$, therefore $dx = \frac{dt}{3}$
Inserted into the integral: $\frac13\int\frac{cost}{sint}dt$, which leaves me with $\frac13\int cot(t)dt$ and then it stops for me, I tried Per-Partes, but couldn't come up with anything useful.
How do I solve these 2 integrals? They're driving me nuts.
Answer
Hints : For the first one use partial fractions, noting that
$$
\frac{1-x}{(x-2)(x+3)}=\frac{-1}{5x-10}-\frac{4/5}{x+3}.
$$
For the second one there is an obvious change of variable.
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