How do you integrate $int 1/((x+7)(x^2+9))$ using partial fractions?

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Answer 1 · 2017-10-02T20:03:12

$int dx/((x+7)(x^2+9)) = 1/58 ln ( abs(x+7)/sqrt(x^2+9)) +7/174arctan(x/3) +C$

Since one of the factors of the denominator is of second degree we apply partial fractions decomposition as:

$1/((x+7)(x^2+9)) = A/(x+7)+ (Bx+C)/(x^2+9)$

$1/((x+7)(x^2+9)) = (A(x^2+9)+ (Bx+C)(x+7))/((x+7)(x^2+9))$

$1 = Ax^2 +9A +Bx^2+Cx+7Bx+7C$

$1= (A+B)x^2 +(7B+C)x +(9A+7C)$

Equation the coefficients of the same degree in $x$ :

${(A+B=0),(7B+C=0),(9A+7C=1):}$

${(A=-B),(7B+C=0),(-9B+7C=1):}$

${(A=-B),(B=-C/7),(9/7C+7C=1):}$

${(A=1/58),(B=-1/58),(C=7/58):}$

So:

$int dx/((x+7)(x^2+9)) = 1/58int dx/(x+7)- 1/58 int x /(x^2+9)dx+7/58 int dx/(x^2+9)$

Solve the three integrals separately:

$int dx/(x+7) = ln abs(x+7) +C_1$

$int x /(x^2+9)dx = 1/2 int (d(x^2+9))/(x^2+9) = 1/2 ln (x^2+9) + C_2$

$int dx/(x^2+9) = 1/3int (d(x/3))/( ( x/3)^2 +1) = 1/3 arctan(x/3) +C_3$

putting the solutions together:

$int dx/((x+7)(x^2+9)) = 1/58 ( ln abs(x+7) -1/2 ln (x^2+9) ) +7/174arctan(x/3) +C$

and finally using the properties of logarithms:

$int dx/((x+7)(x^2+9)) = 1/58 ln ( abs(x+7)/sqrt(x^2+9)) +7/174arctan(x/3) +C$

How do you integrate int 1/((x+7)(x^2+9))int 1/((x+7)(x^2+9)) using partial fractions?