Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 27

int (x^3+x^2+2x+1)/((x^2+1)(x^2+2))dx
To solve, apply partial fractions decomposition.
To express the integrand as sum of proper rational expressions, set the equation as follows:
(x^3+x^2+2x+1)/((x^2+1)(x^2+2)) = (Ax+B)/(x^2+1)+(Cx+D)/(x^2+2)
Multiply both sides by the LCD.
x^3+x^2+2x+1=(Ax+B)(x^2+2)+(Cx+D)(x^2+1)
x^3+x^2+2x+1=Ax^3+2Ax+Bx^2+2B + Cx^3+Cx+Dx^2+D
x^3+x^2+2x+1=(A+C)x^3+(B+D)x^2+(2A+C)x+2B+D
For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the polynomials equal to each other.
x^3:
1=A+C (Let this be EQ1.)
x^2:
1=B+D (Let this be EQ2.)
x:
2=2A+C (Let this be EQ3.)
Constant:
1=2B+D (Let this be EQ4.)
To solve for the values of A, B, C and D, isolate the C in EQ1.
1=A+C
1-A=C
Plug-in this to EQ3.
2=2A+C
2=2A+1-A
2=A+1
1=A
Plug-in the value of A to EQ1.
1=A+C
1=1+C
0=C
Also, isolate the D in EQ2.
1=B+D
1-B=D
Plug-in this to EQ4.
1=2B+D
1=2B+1-B
1=B+1
0=B
And plug-in the value of B to EQ2.
1=B+D
1=0+D
1=D
So the partial fraction decomposition of the integrand is:
(x^3+x^2+2x+1)/((x^2+1)(x^2+2)) = (1x+0)/(x^2+1)+(0x+1)/(x^2+2)=x/(x^2+1)+1/(x^2+2)
Taking the integral of this result to:
int(x^3+x^2+2x+1)/((x^2+1)(x^2+2))dx
=int (x/(x^2+1) + 1/(x^2+2))dx
= int x/(x^2+1)dx + int 1/(x^2+2)dx
For the first integral, apply u-substitution method.
u=x^2+1
du=2xdx
(du)/2=xdx
= int 1/u*(du/2) + int 1/(x^2+2)dx
=1/2 int 1/u du + int 1/(x^2+2)dx
= 1/2ln|u| + 1/sqrt2 tan^(-1) (x/sqrt2)+C
And, substitute back u = x^2+1 .
=1/2ln |x^2+1|+1/sqrt2tan^(-1)(x/sqrt2)+C

Therefore, int (x^3+x^2+2x+1)/((x^2+1)(x^2+2))dx=1/2ln |x^2+1|+1/sqrt2tan^(-1)(x/sqrt2)+C .