Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 11

int_0^1 2/(2x^2 + 3x + 1) dx
sol:
int_0^1 2/(2x^2 + 3x + 1) dx
= 2* int_0^1 1/(2x^2 + 3x + 1) dx
Now Find the partial fractions of 1/(2x^2 + 3x + 1)
1/(2x^2 + 3x + 1) = 1/((2x+1)(x+1))
= A/(2x+1) + B/(x+1)
= (A(x+1)+B(2x+1))/((2x+1)(x+1))
Now Equating the numerators we get
1=(A(x+1)+B(2x+1))
= Ax+2Bx+A+B
= (A+2B)x+ (A+B)
Equating the co -efficients of x and the constants
(A+2B) =0
=> A=-2B
and
1=A+B
1=-2B+B
1=-B => B=-1 so A=2
Then ,
1/((2x+1)(x+1)) = A/(2x+1) + B/(x+1)
= 2/(2x+1) - 1/(x+1)
Now,
int 1/((2x+1)(x+1)) dx = int [2/(2x+1) - 1/(x+1)] dx
= int 2/(2x+1) dx - int 1/(x+1) dx
= ln(2x+1) - ln(x+1) +c
Now Apply the limits 0 to 1we get
int_0^1 1/((2x+1)(x+1)) dx
= [ln(2x+1) - ln(x+1) +c]_0^1
=[ ln(2+1) - ln(1+1) ]-[ ln(0+1) - ln(0+1) ]
=[ ln(3) - ln(2) ]-[ ln(1) - ln(1) ]
as ln(1) =0 so,
= [ ln(3) - ln(2) ]
so,Now
int_0^1 2/((2x+1)(x+1)) dx = 2 *[ ln(3) - ln(2) ]
=ln(9) - ln(4)
=ln(9/4)
is the solution
:)