$\frac{x-1}{x^2+3x+2}\\\frac{x-1}{(x+1)(x+2)}\\\frac{x-1}{(x+1)(x+2)}=\frac{A}{x+2}+\frac{B}{x+1}\\x-1=A(x+1)+B(x+2)\\\\\text{ Let } x=-1 \text{ then } -2=A(0)+B(-1+2) \text{ which means } B=-2\\\\\text{ Instead of choose } x \text{ to be -1, let's now set it equal to -2 to find } A\\\\-3=A(-1)+B(-2+2)\\-3=-A\\A=3\\\\ \text{ So your integral becomes } \\\int \frac{3}{x+2} dx+\int \frac{-2}{x+1} dx\\3 \ln|x+2|-2\ln|x+1|+C\\\\or\\-2\ln|x+1|+3\ln|x+2|+C$

8 0

3 years ago

Find the domain of f(x)=x-9/x squared-81

vagabundo [1.1K]

$f(x)=\dfrac{x-9}{x^2-81}\\\\\text{The domain:}\\\\x^2-81\neq0\qquad\text{add 81 to both sides}\\\\x^2\neq81\to x\neq\pm\sqrt{81}\\\\x\neq-9\ \wedge\ x\neq9\\\\Answer:\ \boxed{\text{All real numbers except -9 and 9}}\to\boxed{x\in\mathbb{R}-\{-9,\ 9\}}$

7 0

3 years ago

When rolling a number cube with sides labeled 1 through 6, what is the probability of rolling a number greater than 2?

vlabodo [156]

That would be 4/6, or 2/3 simplified. As a percent, that would be 66%.

I hope this answer helped you! If you have any further questions or concerns, feel free to ask! :)

4 0

3 years ago

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