§(11-3x/(x^2+2x-3))dx?

To solve this integral, we first need to factor the denominator of the fraction:

x^2 + 2x - 3 = (x + 3)(x - 1)

Now we can rewrite the integral as:

∫(11 - 3x)/(x + 3)(x - 1) dx

Next, we need to perform partial fraction decomposition to break down the fraction into simpler terms. Let's assume that the fraction can be written as:

A/(x + 3) + B/(x - 1)

Multiplying both sides by (x + 3)(x - 1), we get:

11 - 3x = A(x - 1) + B(x + 3)

Expanding and simplifying, we get:

11 - 3x = Ax - A + Bx + 3B

Now we can equate the coefficients of x on both sides:

-3 = A + B
And the constant terms on both sides:

11 = -A + 3B

Solving these two equations simultaneously, we get:

A = -2
B = -1

Now we can rewrite the integral as:

∫(-2/(x + 3) - 1/(x - 1)) dx

Integrating each term separately, we get:

-2∫1/(x + 3) dx - ∫1/(x - 1) dx
= -2ln|x + 3| - ln|x - 1| + C

Therefore, the final solution to the integral ∫(11 - 3x)/(x^2 + 2x - 3) dx is:

-2ln|x + 3| - ln|x - 1| + C

∫(11-3x/(x^2 2x-3))dx

=∫11dx-∫3x/(x^2 2x-3)dx

=11x-∫3x/(x(x+3))dx

=11x-∫3/(x+3)dx

=11x-3ln|x+3|+C

Therefore, the answer is 11x-3ln|x+3|+C.