Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 25

int(4x)/(x^3+x^2+x+1)dx
To solve, apply the partial fraction decomposition.
To do so, factor the denominator.
int(4x)/(x^3+x^2+x+1)dx = int(4x)/((x+1)(x^2+1))dx
Then, express the integrand as sum of proper rational expressions.
(4x)/((x+1)(x^2+1))=A/(x+1)+(Bx+C)/(x^2+1)
Multiply both sides by the LCD.
4x =A(x^2+1)+(Bx+C)(x+1)
4x = Ax^2+A + Bx^2+Bx+Cx+C
4x=(A+B)x^2+(B+C)x + A+C
Express the left side as a polynomial with degree 2.
0x^2+4x+0=(A+B)x^2+Cx+A+C
For the two sides to be equal, the two polynomials should be the same. So set the coefficients of the two polynomials equal to each other.
x^2:
0=A+B (Let this be EQ1.)
x:
4=B+C (Let this be EQ2.)
Constant:
0=A+C (Let this be EQ3.)
To solve for the values of A, B and C, isolate the A in EQ1 and the C in EQ2.
EQ1:
0=A+B
-B=A
EQ2:
4 = B + C
4 - B = C
Plug-in them to EQ3.
EQ3:
0=A+C
0=-B+4-B
0=-2B+4
-4=-2B
2=B
And, plug-in the value of B to EQ1 and EQ2.
EQ1:
0 =A + B
0=A+2
-2=A
EQ2:
4=B+C
4=2+C
2=C
So the partial fraction decomposition of the integrand is:
(4x)/(x^3+x^2+x+1) = -2/(x+1) + (2x+2)/(x^2+1)=-2/(x+1)+(2x)/(x^2+1)+2/(x^2+1)
Then, take the integral of it.
int (4x)/(x^3+x^2+x+1)dx
=int (-2/(x+1)+ (2x)/(x^2+1) + 2/(x^2+1))dx
=int -2/(x+1)dx + int(2x)/(x^2+1)dx + int2/(x^2+1)dx
=-2ln|x+1| + ln|x^2+1|+2tan^(-1)(x) +C

Therefore, int (4x)/(x^3+x^2+x+1)dx=-2ln|x+1| + ln|x^2+1|+2tan^(-1)(x) +C .