`int_0^1 2/(2x^2 + 3x + 1) dx` Evaluate the integral

Textbook Question

Chapter 7, 7.4 - Problem 11 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

1 Answer | Add Yours

kspcr111's profile picture

kspcr111 | In Training Educator

Posted on

`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 

:)

We’ve answered 318,988 questions. We can answer yours, too.

Ask a question