Question

Integrate the following functions with respect to x :

`x^3/((x - 1)(x - 2))`

Answer 1

`int x^3/((x - 1)(x - 2))  "d"x = int ((x^3 - 1) + 1)/((x - 1)(x - 2))  "d"x`

= `int ((x^3 - 1)/((x - 1)(x - 2)) + 1/((x - 1)(x - 2))) "d"x`

= `int (x^3 - 1)/((x - 1)(x - 2))  "d"x + int ("d"x)/((x - 1)(x - 2))`

= `int ((x - 1)(x^2 + x + 1))/((x - 1)(x - 2))  "d"x + int ("d"x)/((x - 1)(x - 2))`

= `int ((x^2 + x + 1))/(x - 2)  "d"x + int ("d"x)/((x - 1)(x - 2))`  ........(1)

Consider `int (x^2 + x + 1)/(x - 2)  "d"x`.

As the degree of the N.R is greater than the degree of the D.R divide the N.R by D.R till the degree of the N.R less than the degree of the D.R.

`(x^2 + x + 1)/(x - 2) = x + 3 + 7/(x - 2)`

`int (x^2 + x + 1)/(x - 2) * "d"x = int [(x + 3) + 7/(x - 2)]  "d"x`

= `int (x + 3) "d"x + int 7/(x - 2)  "d"x`

= `int x  "d"x + 3 int  "d"x + 7 int ("d"x)/(x - 2)`

`int (x^2 + x + 1)/(x - 2) * "d"x = x^2/2 + 3x + 7 log |x - 2|`  ........(2)

Consider `int ("d"x)/((x - 1)(x - 2))`

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

1 = A(x – 2) + B(x – 1)

Put x =

1 = A(2 – 2) + B(2 – 1)

1 = A × 0 + B × 1

B = 1

Put x = 1

1 = A(1 – 2) + B(1 – 1)

1 = A × – 1 + B × 0

A = – 1

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

`int ("d"x)/((x - 1)(x - 2)) = int (- 1/(x - 1) + 1/(x - 2))  "d"x`

= `int - ("d"x)/(x - 1) + int ("d"x)/(x - 2)`

= `- log |x - 1| + log |x - 2| + "c"`  ........(3)

Using equations (2) and (3), equation (1) becomes

`int x^3/((x - 1)(x - 2))  "d"x = x^2/2 + 3x + 7 log |x - 2| - log |x - 1| + log |x - 2| + "c"`

= `x^2/2 + 3x + 8 log |x - 2| - log |x - 1 + "c"`