Question

How do you integrate `int x / [x^2 + 4x + 13]` using partial fractions?

Answer 1

`int x/(x^2+4x+13) d x=(l n(|x^2+4x+13|))/2-2/3 arc tan((x+2)/3)+C`

Explanation:

`int x/(x^2+4x+13) d x=?`

`2/2(x+2-2)=1/2(2x+4-4)`

`int x/(x^2+4x+13) d x=1/2 int(2x+4-4)/(x^2+4x+13) d x`

`int x/(x^2+4x+13) d x=1/2 [color(red)(int (2x+4)/(x^2+4x+13)d x)-4color(green)( int (d x)/(x^2+4x+13))]`

`" solve the integration ;"`

`color(red)(int (2x+4)/(x^2+4x+13)d x)`

`"substitute "u=x^2+4x+13" ; " d u=2x+4`

`color(red)(int (2x+4)/(x^2+4x+13)d x)=int (d u)/u=l n u`

`"undo substitution "`

`color(red)(int (2x+4)/(x^2+4x+13)d x)=l n(x^2+4x+13)`

`"now solve ;"`

`color(green)(-4 int (d x)/(x^2+4x+13))=-4 int (d x)/(x^2+4x+4+9)`

`color(green)(-4 int (d x)/(x^2+4x+13))=-4 int(d x)/((x+2)^2+3^2)`

`"substitute "`

`u=x+2" ; " d u= d x`

`color(green)(-4 int (d x)/(x^2+4x+13))=-4 int (d u)/(u^2+3^2)=-4/3 arc tan (u/3)`

`"undo substitution "`

`color(green)(-4 int (d x)/(x^2+4x+13))=-4/3 arc tan ((x+2)/3)`

`"Integration have solved"`

`int x/(x^2+4x+13) d x=1/2(l n(x^2+4x+13)-4/3 arc tan((x+2)/3))`

`int x/(x^2+4x+13) d x=(l n(|x^2+4x+13|))/2-2/3 arc tan((x+2)/3)+C`