$\frac{x^3+2x^2-x-2}{(x-1)} =x^2+3x+2 \\ \\ x^3+2x^2-x-2= \\= x^2(x+2)-(x+2) \\ =(x+2)(x^2-1)=(x+2)(x-1)(x+1) \\ \\ \frac{x^3+2x^2-x-2}{(x-1)} = \frac{(x+2)(x-1)(x+1)}{x-1} =(x+2)(x+1)=x^2+3x+2$

7 0

3 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Rus_ich [418]

Answer:

Diverges

Step-by-step explanation:

We can solve this by using integral by parts:

Let

$u=lnx$ $dv=(1/x)dx$

$du=(1/x)dx$ $v=lnx$

$\int\limits {lnx/x} \, dx = (lnx)^2-\int\limits {lnx/x} \, dx$

We can add

$\int\limits {lnx/x} \, dx$ to both sides

$\int\limits{lnx/x} \, dx = (1/2)\cdot{(lnx)^2}$

We can evaluate the limit between 2 and infinity.

If x tends to infinity the limit will be infinity and therefore the integral diverges to ∞

6 0

3 years ago