What is integral 1/(x-1)(x+2)?

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You need to evaluate the indefinite intregral, such that:

`int (dx)/((x-1)(x+2))`

You need to use the partial fraction expansion to split the given integral into two simpler integrals, such that:

`1/((x-1)(x+2)) = a/(x-1) + b/(x+2)`

`1 = a(x+2) + b(x-1) `

`1 = ax + 2a + bx - b`

`1 = x(a+b) + 2a - b`

Equating the coefficients of like powers yields:

`{(a + b = 0),(2a - b = 1):} => {(a = -b),(-2b - b = 1):}`

`{(a = -b),(-3b = 1):} => {(a = -b),(b = -1/3):} => {(a = 1/3),(b = -1/3):}`

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

Integrating both sides yields:

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

Using the property of linearity of intgeral, yields:

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

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

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

**Hence, evaluating the indefinite integral, using partial fraction expansion, yields **`int 1/((x-1)(x+2)) dx = (1/3)(ln|(x-1)/(x+2)|) + c.`

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