Question

Integrate the rational function:
5x/ (x + 1)(x2 - 4)​

Answer 1

Answer :

(5/3)·log(x+1) – (5/2)·log(x+2) + (5/6)·log(x-2) + c

Solution :

Here ,

We need to integrate the given rational function 5x/(x+1)(x²-4) with respect to x .

Firstly ,

Let's further factorize the denominator of the given rational function .

=> Denominator = (x + 1)(x² - 4)

=> Denominator = (x + 1)(x² - 2²)

=> Denominator = (x + 1)(x + 2)(x - 2)

Thus ,

The given rational function can be rewritten as ; 5x/(x + 1)(x + 2)(x - 2)

Now ,

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

=> 5x/(x+1)(x+2)(x-2) = [ A(x+2)(x-2) + B(x+1)(x-2) + C(x+1)(x+2) ] / (x + 1)(x + 2)(x - 2)

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

• If x = -1 , then

=> 5•(-1) = A(-1+2)(-1-2) + B(-1+1)(-1-2) + C(-1+1)(-1+2)

=> -5 = A•1•(-3) + B•0•(-3) + C•0•1

=> -5 = -3A + 0 + 0

=> -3A = -5

=> A = -5/-3

=> A = 5/3

• If x = -2 , then

=> 5•(-2) = A(-2+2)(-2-2) + B(-2+1)(-2-2) + C(-2+1)(-2+2)

=> -10 = A•0•(-4) + B•(-1)•(-4) + C•1•0

=> -10 = 0 + 4B + 0

=> 4B = -10

=> B = -10/4

=> B = -5/2

• If x = 2 , then

=> 5•2 = A(2+2)(2-2) + B(2+1)(2-2) + C(2+1)(2+2)

=> 10 = A•4•0 + B•3•0 + C•3•4

=> 10 = 0 + 0 + 12C

=> 12C = 10

=> C = 10/12

=> C = 5/6

Now ,

=> 5x/(x + 1)(x² - 4)

= 5x/(x+1)(x+2)(x-2)

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

= (5/3)/(x+1) + (-5/2)/(x+2) + (5/6)/(x-2)

Now ,

Integrating with respect to x , we get ;

→ (5/3)·log(x+1) - (5/2)·log(x+2) + (5/6)·log(x-2) + c , where c is the integrating constant

Hence ,

Required answer is : (5/3)·log(x+1) – (5/2)·log(x+2) + (5/6)·log(x-2) + c .