To evaluate the integral \( \int\frac{2x+3}{x^3+x^2-2x}dx \), we will use the method of partial fraction decomposition. This method is essential for integrating rational functions where the degree of the numerator is less than the degree of the denominator.

Factoring the Denominator

First, we need to factor the denominator completely. The denominator is a cubic polynomial: \( x^3+x^2-2x \).

• Factor out the common term \( x \):
  \( x^3+x^2-2x = x(x^2+x-2) \)
• Factor the quadratic expression \( x^2+x-2 \). We look for two numbers that multiply to -2 and add to 1 (which are 2 and -1):
  \( x^2+x-2 = (x+2)(x-1) \)

So, the fully factored denominator is \( x(x+2)(x-1) \). This factorization is crucial for the partial fraction decomposition.

Partial Fraction Decomposition Setup

Now, we can set up the partial fraction decomposition for the integrand \( \frac{2x+3}{x(x+2)(x-1)} \). Since all factors in the denominator are distinct linear factors, the decomposition will be of the form:

To find the constants A, B, and C, we multiply both sides of the equation by the common denominator \( x(x+2)(x-1) \):

Determining the Constants (A, B, C)

We can find the values of A, B, and C by substituting specific values of \( x \) that make some terms zero, simplifying the calculation of each constant.

• For constant \( A \): Set \( x=0 \)
  Substitute \( x=0 \) into the equation \( 2x+3 = A(x+2)(x-1) + Bx(x-1) + Cx(x+2) \): \[ 2(0)+3 = A(0+2)(0-1) + B(0)(0-1) + C(0)(0+2) \] \[ 3 = A(2)(-1) + 0 + 0 \] \[ 3 = -2A \] \[ A = -\frac{3}{2} \]
• For constant \( C \): Set \( x=1 \)
  Substitute \( x=1 \) into the equation: \[ 2(1)+3 = A(1+2)(1-1) + B(1)(1-1) + C(1)(1+2) \] \[ 5 = A(3)(0) + B(1)(0) + C(1)(3) \] \[ 5 = 0 + 0 + 3C \] \[ 5 = 3C \] \[ C = \frac{5}{3} \]
• For constant \( B \): Set \( x=-2 \)
  Substitute \( x=-2 \) into the equation: \[ 2(-2)+3 = A(-2+2)(-2-1) + B(-2)(-2-1) + C(-2)(-2+2) \] \[ -4+3 = A(0)(-3) + B(-2)(-3) + C(-2)(0) \] \[ -1 = 0 + 6B + 0 \] \[ -1 = 6B \] \[ B = -\frac{1}{6} \]

Integrating the Partial Fractions

Now that we have the values of A, B, and C, we can rewrite the original integral using the partial fraction decomposition:

We can split this into three simpler integrals, each involving a basic logarithmic form:

Each of these integrals is of the form \( \int \frac{1}{u} du = \log|u| + c \). So, we integrate term by term:

• The integral of \( \frac{1}{x} \) is \( \log|x| \).
• The integral of \( \frac{1}{x+2} \) is \( \log|x+2| \).
• The integral of \( \frac{1}{x-1} \) is \( \log|x-1| \).

Substituting these back into our expression, and adding the constant of integration \( c \):

Rearranging the terms to match the format of the options, it is customary to put the positive terms first for clarity:

Final Integral Result

The result of the integration is \( \frac{5}{3}\log|x-1| - \frac{3}{2}\log|x| - \frac{1}{6}\log|x+2| + c \). This comprehensive solution demonstrates the application of partial fractions for rational function integration.