Indefinite integral

Question

∫(5x+1)/((x-4)*(x+1)2) dx

Answer 1

The way to solve this integral is partial fraction decomposition, thus you can write:

(5x+1)/[(x-4)*(x+1)²] = A/(x-4) + B/(x+1) + C/(x+1)² . When you combine for a common denominator, you get:

[A(x+1)² + B(x-4)(x+1) + C(x-4)] / [(x-4)*(x+1)²]. Notice that then you have:

Ax² + 2Ax + A + Bx² - 3Bx - 4B + Cx - 4C = 5x + 1.

=> (A + B)x² + (2A - 3B + C)x + (A - 4B - 4C) = 5x + 1.

=> A + B = 0

   2A - 3B + C = 5

     A - 4B - 4C = 1

 

Sub A = -B

=> -5B + C = 5

     -5B - 4C = 1

 

=> 5C = 4

=> C = 4/5

 

=> B = -21/25

 

=> A = 21/25

 

Thus we can rewrite the integral as:

 

21/25*∫ [1/(x-4)]dx - 21/25*∫ [1/(x+1)]dx + 4/5*∫ [1/(x+1)²]dx

 

From this we know:

 

21/25*ln(x-4) - 21/25*ln(x+1) - 4/5*[1/(x+1)] + C.

21/25*ln[(x-4)/(x+1)] - 4/5*[1/(x+1)] + C.