Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 26

int (x^2+x+1)/(x^2+1)^2 dx
To solve, apply partial fraction decomposition.
To express the integrand as sum of proper rational functions, set the equation as follows:
(x^2+x+1)/(x^2+1)^2=(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2
Multiply both sides by the LCD.
x^2+x+1=(Ax+B)(x^2+1) + Cx + D
x^2+x+1=Ax^3+Bx^2+Ax+B+Cx+D
x^2+x+1=Ax^3+Bx^2+(A+C)x+B+D
Express the left side as a polynomial with degree 3.
0x^3+x^2+x+1=Ax^3+Bx^2+(A+C)x+B+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:
0=A (Let this be EQ1.)
x^2:
1=B (Let this be EQ2.)
x:
1=A+C (Let this be EQ3.)
Constant:
1=B+D (Let this be EQ4.)
In the equations, the values of A and B are already known. So only the values of C and D have to be solved. To do so, plug-in the value of A to EQ3.
1=A+C
1=0+C
1=C
Also, plug-in the value of B to EQ4.
1=B+D
1=1+D
0=D
So the partial fraction decomposition of the integrand is:
(x^2+x+1)/(x^2+1)^2=(0x+1)/(x^2+1)+(1x+0)/(x^2+1)^2=1/(x^2+1)+x/(x^2+1)^2
Taking the integral of it result to:
int (x^2+x+1)/(x^2+1)^2 dx
= int (1/(x^2+1)+x/(x^2+1)^2) dx
= int 1/(x^2+1)dx + int x/(x^2+1)^2dx
For the first integral, apply the formula int 1/(u^+a^2)=1/a tan^(-1) u/a+C.
For the second integral, apply u-substitution method.
u=x^2+1
du=2x dx
(du)/2=xdx
= int 1/(x^2+1)dx + int 1/u^2 * (du)/2
= int 1/(x^2+1)dx +1/2 int u^(-2) du
= tan^(-1)x -1/2u^(-1)+C
= tan ^(-1)x - 1/(2u)+C
Substitute back u=x^2+1 .
= tan ^(-1)x - 1/(2(x^2+1))+C

Therefore, int (x^2+x+1)/(x^2+1)^2dx= tan ^(-1)x - 1/(2(x^2+1))+C .