91Ó°ÊÓ

To integrate the given function, we first need to factor the denominator. The denominator is a quadratic function: $$ x^2 + 2x - 3 $$ We can factor this as: $$ (x+3)(x-1) $$ So our integral now looks like the following: $$ \int \frac{3x+5}{(x+3)(x-1)}dx $$

Now we need to perform partial fraction decomposition to break down the fraction into simpler terms that are easier to integrate. This involves turning the fraction into the sum of two simpler fractions: $$ \frac{3x+5}{(x+3)(x-1)} = \frac{A}{x+3} + \frac{B}{x-1} $$ We will now solve for A and B. To do this, we can first write the fractions on the right side using a common denominator: $$ \frac{A(x-1) + B(x+3)}{(x+3)(x-1)} $$ Now, the numerators must be the same for both sides to be equal: $$ 3x + 5 = A(x-1) + B(x+3) $$ Now, we can pick two convenient values for x that will make the other terms equal to zero: Let x = 1: $$ 3(1) + 5 = A(1 - 1) + B(1 + 3) $$ $$ 8 = 4B $$ So, B = 2. Let x = -3: $$ 3(-3) + 5 = A(-3-1) + B(-3+1) $$ $$ -4 = -4A $$ So, A = 1. Our partial fraction decomposition is now: $$ \frac{3x+5}{(x+3)(x-1)} = \frac{1}{x+3} + \frac{2}{x-1} $$

Now we integrate each term of the partial fraction decomposition separately: $$ \int \left(\frac{1}{x+3} + \frac{2}{x-1}\right)dx $$ $$ = \int \frac{1}{x+3}dx + \int\frac{2}{x-1}dx $$ $$ =\ln |x+3| + 2\ln |x-1| + C $$ So the result of the integral is: $$ \int \frac{3x+5}{x^{2}+2x-3}dx = \ln |x+3| + 2\ln |x-1| + C $$