Today's math problems — Systems of Linear Equations, Integrals, Algebra & more

$\bullet$ Decompose by partial fractions:

$${\displaystyle \frac{1}{x^2-1}}={\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{x-1}}-{\displaystyle \frac{1}{x+1}})$$

$\bullet$ Integrate term-by-term:

$$\int {\displaystyle \frac{dx}{x^2-1}}={\displaystyle \frac{1}{2}}(\ln \mathrm{\mid }x-1\mathrm{\mid }-\ln \mathrm{\mid }x+1\mathrm{\mid })+C$$

$\bullet$ Combine logs:

$$={\displaystyle \frac{1}{2}}\ln \mid {\displaystyle \frac{x-1}{x+1}}\mid +C$$

Associative ✓, identity $0$ ✓, inverse of $n$ is $-n$ ✓, commutative ✓. So $(\mathbb{Z},+)$ is an abelian group.

The three edges from origin are $(3,0,0),(0,4,0),(0,0,5)$. Determinant: $3\cdot 4\cdot 5=60$. Volume: $60\mathrm{/}6=10$.

$f^{\mathrm{\prime }}(x)=3x^2-3$ vanishes at $x=\pm 1$. $f(-1)=-1+3+1=3>0$ and $f(1)=1-3+1=-1<0$. The local max is positive and the local min is negative, so the cubic crosses the $x$-axis three times.

Set the terms as $5-r,5,5+r$. Then $(5-r{)}^2+25+(5+r{)}^2=50+2r^2+25=83$, so $2r^2=8\Rightarrow r=2$. The largest term is $5+2=7$.

The tetrahedron with these vertices has volume $1\mathrm{/}6\mathrm{\ne }0$, so the four points are NOT coplanar — they form a non-degenerate tetrahedron.