# Explain integral (2x+3)/x(x-1)(x+2)=?

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You need to use a procedure called partial fraction decomposition to solve the integral as a sum of simpler integrals.

The fraction `(2x+3)/(x(x-1)(x+2))` may be decomposed in the form:

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

Clear the coefficients:

`2x + 3 = A(x-1)(x+2) + Bx(x+2) + Cx(x-1)`

Opening the brackets yields:

`2x + 3 = Ax^2 + Ax - 2A + Bx^2 + 2Bx + Cx^2 - Cx`

Equate the coefficients of the same powers:

A+B+C = 0

A+2B-C = 2

`-2A = 3 =gt A = -3/2`

Substitute `A = -3/2:`

`` `B+C = 3/2`

`` `2B-C = 2 + 3/2`

`` Adding the equations yields `3B = 5 =gt B = 5/3`

`C = 3/2 - 5/3 =gt C = (9-10)/6 = -1/6`

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

Integrating both sides yields:

`int ((2x+3)dx)/(x(x-1)(x+2)) = -int (3dx)/(2x) + int (5dx)/(3(x-1))- intdx/(6(x+2))`

`int ((2x+3)dx)/(x(x-1)(x+2)) = (-3/2)*ln |x| + (5/3)*ln |x-1| - (1/6)*ln|x+2| + c`

**Solving the integral yields `int ((2x+3)dx)/(x(x-1)(x+2)) = (-3/2)*ln |x| + (5/3)*ln |x-1| - (1/6)*ln|x+2| + c` **