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

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Answer 1 · 2016-08-10T17:36:18

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

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

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

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

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

$" solve the integration ;"$

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

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

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

$"undo substitution "$

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

$"now solve ;"$

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

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

$"substitute "$

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

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

$"undo substitution "$

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

$"Integration have solved"$

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

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

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