Indefinite integral

Question

∫(5x+1)/(x-4)(x+1)² dx

Answer 1

This is an integration by partial fraction decomposition problem.

We have a polynomial over a polynomial. Degree of numerator is less than degree of denominator so no long division is needed.

Next we need to factor denominator completely (not needed as that's already done in this case).

Now we setup our partial fractions. We have 1 DLF (distinct linear factor (x-4)) and 1 RLF (repeated linear factor (x+1)²), which means we need 3 total partial fractions based on adding multiplicity of our factors.

5x+1 / (x-4)(x+1)² = A/ (x-4) + B/ (x+1) + C/ (x+1)²

5x + 1 = A(x+1)² + B (x+1)(x-4) + C(x-4)

Now we can use various method to solve for our variables. I will avoid distributing and using systems of equations by choosing x values that make my factors go to 0.

So, let's pick x=4 first. We get 5(4) + 1 = A(4+1)²

20 +1= A(5)²

21= 25A

A= 21/25

Next, let's pick x= -1 We get 5(-1) + 1 = C(-1-4)

-5 + 1 = C(-5)

-4 = -5C

C = 4/5

Lastly, we choose any other number to get B. I'll go with x=0. We will need to substitute the known values for A and C here to get B.

5(0) + 1 = A(0+1)² + B (0+1)(0-4) + C(0-4)

1 = 21/25(1)² + B (1)(-4) + 4/5(-4)

1 = 21/25 + -4B - 16/5

1+4B = (21 - 16*5) / 5

1+4B = (21 - 80) / 5

1+4B = (-59) / 5

4B = -59/5 - 1

4B = - 64/5

B = - 64/5 * 1/4

B = -16/5

Now we have our variables, let's construct our new expression and integrand with it.

21/25 / (x-4) + -16/5 / (x+1) + 4/5 / (x+1)²

∫ 21/25 / (x-4) + -16/5 / (x+1) + 4/5 / (x+1)² dx

= 21/25 ln | x - 4 | - 16/5 ln |x+1| - 4/ (5(x+1)) + C <-- we can use log properties to simplify a bit more but it's not necessary.