Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 13

int (4x^2+2x-1)/(x^3+x^2)dx
To solve using partial fraction method, the denominator of the integrand should be factored.
(4x^2+2x-1)/(x^3+x^2)=(4x^2+2x-1)/(x^2(x+1))
Take note that if the factor in the denominator is linear and non-repeating, each factor in the denominator has a partial fraction form of A/(ax+b) .
And if the factor is linear and repeating, its partial fraction decomposition has a form A_1/(ax+b) + A_2/(ax+b)^2+... +A_n/(ax+b)^n .
So, expressing the integrand as sum of fractions, it becomes:
(4x^2+2x-1)/(x^2(x+1))=A/x+B/x^2+C/(x+1)
To determine the values of A, B and C, multiply both sides by the LCD of the fractions present.
x^2(x+1)*(4x^2+2x-1)/(x^2(x+1))=(A/x+B/x^2+C/(x+1))*x^2(x+1)
4x^2+2x-1=Ax(x+1)+B(x+1)+Cx^2
Then, assign values to x in which either x, x^2 or x+1 will become zero.
So plug-in x=0 to get the value of B.
4(0)^2+2(0)-1=A(0)(0+1)+B(0+1)+C(0)^2
-1=A(0)+ B(1)+C(0)
-1=B
Also, plug-in x=-1 to get the value of C.
4(-1)^2+2(-1)-1=A(-1)(-1+1)+B(-1+1)+C(-1)^2
1=A(0)+B(0)+C(1)
1=C
To get the value of A, plug-in the values of B and C. Also, assign any value to x. Let it be x=1.
4(1)^2+2(1)-1=A(1)(1+1)+ (-1)(1+1)+1(1)^2
5=A(2)-2+1
5=2A-1
6=2A
3=A
So, the partial fraction decomposition of the integrand is:
int(4x^2+2x-1)/(x^3+x^2)dx
= int (4x^2+2x-1)/(x^2(x+1))dx
= int (3/x -1/x^2+1/(x+1))dx
Then, express it as three integrals.
= int 3/x dx - int 1/x^2 dx + int 1/(x+1)dx
= 3int 1/x dx- int x^(-2) dx + int 1/(x+1)dx
For the first and third integral, apply the formula int 1/u du = ln|u|+C .
And for the second integral, apply the formula int u^n du = u^(n+1)/(n+1)+C .
= 3ln|x| + x^(-1) + ln|x+1| +C
=3ln|x| +1/x + ln|x+1| +C

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

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