To solve the inequality:

log_3 ((2x^2+3x-5)/(x+1)) <= 1

First, rewrite the inequality in exponential form:
3^(log_3 ((2x^2+3x-5)/(x+1))) <= 3^1

This simplifies to:
(2x^2+3x-5)/(x+1) <= 3

Next, multiply both sides by (x+1) to get rid of the denominator:
2x^2 + 3x - 5 <= 3x + 3

Subtract 3x and 3 from both sides:
2x^2 - 3x - 8 <= 0

Now, factor the quadratic equation:
(2x + 1)(x - 8) <= 0

This inequality holds true when either both factors are positive or both are negative. So we have two cases to consider:

Case 1: 2x + 1 > 0 and x - 8 > 0
This occurs when x > -1/2 and x > 8, which simplifies to x > 8.

Case 2: 2x + 1 < 0 and x - 8 < 0
This occurs when x < -1/2 and x < 8, which simplifies to x < -1/2.

Therefore, the solution to the inequality log_3 ((2x^2+3x-5)/(x+1)) <= 1 is x < -1/2 or x > 8.