Question

Find the integral of the function 1/x^2+3x+1 dx

Answer 1

Step-by-step explanation:

We have,

$\int \frac{1 \: \: dx}{ {x}^{2} + 3x + 1}$

$= \int \frac{dx}{(x + 2)(x + 1)}$

$= \int \frac{((x + 2) - (x + 1))dx}{(x + 2)(x + 1)}$

$= \int \frac{dx}{(x + 1)} - \int \frac{dx}{(x + 2)}$

$= ln |x + 1| - ln |x + 2| + c$

$= ln | \frac{x + 1}{x + 2} | + c$