# Determine the antiderivative of y=1/(x-1)(x+4).

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To determine the indefinite integral, we'll write the function as a sum or difference of elementary ratios.

1/(x-1)(x+4) = A/(x-1) + B/(x+4)

1 = A(x+4) + B(x-1)

We'll remove the brackets:

1 = Ax + 4A + Bx - B

We'll combine like terms:

1 = x(A+B) + 4A - B

A + B = 0

A = -B

4A - B = 1

4A + A = 1

5A = 1

A = 1/5

B = -1/5

1/(x-1)(x+4) = 1/5(x-1) - 1/5(x+4)

Int dx/(x-1)(x+4) = Int dx/5(x-1) - Int dx/5(x+4)

Int dx/(x-1)(x+4) = (1/5) (ln |x-1| - ln|x+4|) + C

**Int dx/(x-1)(x+4) = (1/5)ln |(x-1)/(x+4)| + C**

To determine the antiderivative of y = 1/(x-1)(x+4).

To determine th antiderivative.

We first split 1/(x-1)(x+4) into partial fractions .

1/((x-1)(x+4) = A/(x-1)+B/(x+4)

We multiply both sides by (x-1)(x+4).

1 = A(x+4)+B(x-1)....(1)

Put x= 1.Then 1 = A(1+4)+B*0.

1=5A. So A = 1/5.

Purt x= -4 in (1) : Then 1 = A*0 +B(-4 -1) = -5B.

Therefore 1 = -5B. So B = -1/5.

Therefore 1/(x-1)(x+1) = 1/5(x-1) -1/5(x+1).

We now find the antiderivative of 1/(x-1)(x+4).

Int dx/(x-1)(x+1) = Int {dx/5(x-1)-dx/5(x+1)}+C.

Int dx/(x-1)(x+1) = Int dx/5(x-1) - Int dx/(5(x+1)+C.

Int dx/(x-1)(x+1) = (1/5) log (x-1) - (1/5) log (x+1)+C.

Int dx/(x-10(x+1) = (1/5) { log(x-1) - og(x+1)}+C.

Int dx/(x-1)(x+1) = (1/5)log (x-1)/(x+1)+C

Therefore, the antiderivative of dx/(x-1)(x+1) is (1/5)log (x-1)/(x+1)+C.

We have to find the integral of y=1/(x-1)(x+4)

We know that the integral of 1/(1+x) ln (1+x). We need to convert the given expression to a sum of two terms with a similar form.

Let's write 1/(x-1)(x+4) as P/(x-1) + Q/(x+4)

Now we have to determine P and Q.

1/(x-1)(x+4) = P/(x-1) + Q/(x+4)

multiply all terms by (x-1)(x+4)

=> 1 = P*(x+4) + Q*(x-1)

=> 1= Px + 4P + Qx - Q

=> 1 = x(P+Q) + 4P - Q

=> P + Q = 0 adn 4P - Q = 1

P+ Q = 0

=> P = -Q

Putting this in 4P - Q = 1

=> -4Q - Q = 1

=> Q = -1/5

And P = 1/5

1/(x-1)(x+4) = 1/5(x-1) - 1/5(x+4)

Now Int [ 1/5(x-1) - 1/5(x+4)]

= Int [1/5(x-1)] - Int [1/5(x+4)]

= (1/5) ln (x-1) - (1/5) ln ( x+4)

=> (1/5) ln [ (x -1)/ (x+4)] + C

**Therefore integral of 1/(x-1)(x+4) is (1/5) ln [ (x -1)/ (x+4)] + C.**