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)(x2+9)=Ax+7+Bx+Cx2+9
1/((x+7)(x^2+9)) = (A(x^2+9)+ (Bx+C)(x+7))/((x+7)(x^2+9))1(x+7)(x2+9)=A(x2+9)+(Bx+C)(x+7)(x+7)(x2+9)
1 = Ax^2 +9A +Bx^2+Cx+7Bx+7C1=Ax2+9A+Bx2+Cx+7Bx+7C
1= (A+B)x^2 +(7B+C)x +(9A+7C)1=(A+B)x2+(7B+C)x+(9A+7C)
Equation the coefficients of the same degree in xx:
{(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
