how to you use partial fractions to compute the integral of ax/((x^2)-bx)dx?

a x dx /[x (x-b)] ???? but divide top and bottom by x, do not expand in

a/x + d/(x-b)
strange

To compute the integral of ax/((x^2)-bx)dx, you can use partial fractions. Here's how to do it:

1. Factor the denominator: (x^2) - bx = x(x - b/x)

2. Write the given expression as a sum of fractions with simpler denominators: ax/((x^2)-bx)dx = A/x + B/(x - b/x)

3. Multiply both sides of the equation by the denominator to get rid of the fractions: ax = A(x - b/x) + B(x)

4. Simplify the equation: ax = A(x^2 - b) + B(x^2)

5. Expand: ax = Ax^2 - Ab + Bx^2

6. Combine like terms: (A + B) x^2 - Ab = ax

7. Equate the coefficients of corresponding powers of x on both sides of the equation.

For the constant term, A + B = 0 => A = -B.
For the coefficient of x, -Ab = a => B = -a/b.

Therefore, A = a/b and B = -a/b.

8. Rewrite the original integral using the partial fractions: ax/((x^2)-bx)dx = (a/b) * (1/x) + (-a/b) * (1/(x - b/x))dx

9. Integrate each term separately.

- Integral of (a/b)*(1/x)dx is (a/b) * ln|x| + C, where C is the constant of integration.

- Integral of (-a/b)*(1/(x - b/x))dx requires a substitution. Let u = x - b/x, then du = (1 + b/x^2)dx.
Rearranging gives dx = (x^2 / (x^2 + bx))du.
Substituting these values into the integral, it becomes:

Integral of (-a/b) * (x^2 / (x^2 + bx))du.

To simplify further, rewrite x^2 as (x*(x^2))/(x^2). This allows us to cancel out a factor of x in the numerator and denominator, giving:

Integral of (-a/b) * (x / (x + b))du.

Now, you can integrate (-a/b) * (x / (x + b))du. The result will be (-a/b) * ln(x + b) + C, where C is the constant of integration.

10. Finally, add the integrals obtained from step 9 to get the complete solution:

Integral of ax/((x^2)-bx)dx = (a/b) * ln|x| + (-a/b) * ln(x + b) + C, where C is the constant of integration.