Find the exact value of the following giving your answer in the form k + In a where K is an interger and In a is a simplified fraction

Question

∫^1/2     11x - 1         dx

0     (1- x)²(2 + 3x)

Answer 1

1) transform the expression under the integral sign into the following sum:

 

11x-1           =A/(2+3x)+(Bx+C)/(x-1)²;

(1-x)²(2+3x)

 

Undetermined coefficients A, B, and C can be found by bringing the right side to common denominator and equating numerators on the left and right side. You will get:

 

A(x-1)²+(2+3x)(Bx+C)=11x-1

The following system of equations results:

 

A+3B=0;

2B-2A+3C=11;

A+2C=-1.

 

I will leave to you to solve it, giving the result:

 

A=-3, B=1, C=1. So you will get:

 

-3/(2+3x)+(x+1)/(x-1)²=-3/(2+3x)+1/(x-1)+2/(x-1)²;

Now your integral becomes the sum of three integrals:

 

∫^1/2₀(-3*dx/(2+3x)+∫^1/2₀ dx/(x-1)+∫^1/2₀ 2*dx/(x-1)²; 

 

All those integrals can be computed using table integrals; The first two are just logarithms and the third one is the power function. The answer is:

 

[-ln|2+3x|+ln|x-1|-2/(x-1)] |₀^½=[ln|(x-1)/(2+3x)|-2/(x-1)] |₀^½=ln(1/7)-ln(1/2)+4-2=2+ln(2/7)

 

Answer: 2+ln(2/7)