Which is the integral of the differential dx/(x+1)(2x-3)?

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First, we have to write the differential as a sum of simple ratio:

1/(x+1)(2x-3) = A/(x+1) + B/(2x-3)

Bringing the 2 ratio to the same denominator, we'll have:

1 = A(2x-3) + B(x+1)

1 = 2Ax - 3A + Bx + B

1 = x(2A+B) - 3A + B

Being an identity, the expression above has to have the corresponding coefficients equal.

So, the coefficient of x, from the left side, is 0, the left side being written: 0x+1.

According to this:

2A+B=0, so B=-2A

-3A + B=1, so -3A-2A=1

-5A=1

**A=-1/5**

B=-2A=-2(-1/5)

**B=2/5**

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

Integral [dx/(x+1)(2x-3)]=Integral -dx/5(x+1)+2Integral [dx/5(2x-3)]

Integral -dx/5(x+1) = -(1/5)ln(x+1)

2Integral [dx/5(2x-3)] = (2/5)ln(2x-3)

**Integral [dx/(x+1)(2x-3)]= (1/5)(ln[(2x-3)^2/(x+1)] + C**

To find the integral of the differential dx/(x+1)(2x-3)"

Before integrating the rational algebraic expression1/(x+1)(2x-3), we shall resolve it into the sum partial fractions A/(x+1) and B/(x+3)

So 1/[(x+1)(2x-3)] = A/(x+1)+B/(2x-3). Or Multiplyung both sides by (x+1) and (x+3), we get:

1 = A(2x-3)+B(x+1)............(1)

Putting x=3/2, or 2x-3 =0 in .Eq(1) we get: 1 = A*0+b(3/2+1) or B = 2/5.

Again Putting x=-1 Or x+1 = 0 in eq (1) we get:

1 = A*(2*1-3)+B*0. Or 1=-A. Or A =-1.

So the given expression 1/[(x+1)(2x-3)] = -1/(x+1) +2/5)/(2x-3).

Therefore, integral {1/[(x+1)(2x-3)]}dx = Integral [-1/(x+1)]dx - integral (2/5)/(2x-3)+C , where C is the constant of integration.

= ln(x+1) -(2/5)[ln(2x-3)](1/2)+C

=ln(x+1)-(1/5)ln(2x-3) = ln {(x+1)/(2x-3)^(2/5)}+C

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