How do you find $int (3-x)/((x^2+3)(x+3)) dx$ using partial fractions?

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Answer 1 · 2016-04-03T17:10:49

$int (3-x)/((x^2+3)(x+3))dx$
$=-1/4ln (x^2+3)+(sqrt3)/6 tan^-1 ((sqrt3*x)/3)+1/2ln(x+3)+C_0$

We start with partial fraction procedure with the unknowns A, B, C

$(3-x)/((x^2+3)(x+3))=(Ax+B)/(x^2+3)+C/(x+3)$

LCD $=(x^2+3)(x+3)$

$(3-x)/((x^2+3)(x+3))=(Ax^2+Bx+3Ax+3B+Cx^2+3C)/((x^2+3)(x+3))$

Transform the numerator so that the numerical coefficients match for both left and right side of the equation

$(0*x^2+(-1)x+3x^0)/((x^2+3)(x+3))=(Ax^2+Cx^2+3Ax+Bx+3B+3C)/((x^2+3)(x+3))$

$(0*x^2+(-1)x+3x^0)/((x^2+3)(x+3))=((A+C)x^2+(3A+B)x+(3B+3C)x^0)/((x^2+3)(x+3))$

We now have the equations to solve for the values of A,B, C

$A+C=0$
$3A+B=-1$
$3B+3C=3$

Simultaneous solution of these equations result to
$A=-1/2$ and $B=1/2$ and $C=1/2$

$(3-x)/((x^2+3)(x+3))=(-1/2x+1/2)/(x^2+3)+(1/2)/(x+3)$

We can now integrate

$int(3-x)/((x^2+3)(x+3))dx=int [(-1/2x+1/2)/(x^2+3)+(1/2)/(x+3)] dx$

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

$int(3-x)/((x^2+3)(x+3))dx=-1/4ln (x^2+3)+(sqrt3)/6 tan^-1 ((sqrt3*x)/3)+1/2ln(x+3)+C_0$

God bless....I hope the explanation is useful.

How do you find int (3-x)/((x^2+3)(x+3)) dxint (3-x)/((x^2+3)(x+3)) dx using partial fractions?