`int y/((y+4)(2y-1)) dy` Evaluate the integral

Textbook Question

Chapter 7, 7.4 - Problem 10 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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kspcr111 | In Training Educator

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`inty/((y+4)(2y-1)) dy`

sol:-

`inty/((y+4)(2y-1)) dy`

First take partial fractions of `y/((y+4)(2y-1))`

`y/((y+4)(2y-1)) ` = `(A/(y+4)) +(B/(2y-1))`

                        = `((A*(2y-1))+(B*(y+4)))/((y+4)(2y-1))`

     Now equate the numerators we get

`y=((A*(2y-1))+(B*(y+4)))`

  = `2Ay-A+By+4B`

  = `(2A+B)y + (4B-A)`

Now equating the co efficients of y and the constants we get 

`2A+B=1` , `4B-A=0`

=> `4B=A`

`2A+B=1`

=> `2(4B) +B=1`

=> `8B+B=1`

`B=1/9`

so `A= 4/9`

Then 

`y/((y+4)(2y-1)) ` = `(A/(y+4)) +(B/(2y-1))`

                       = `(4/(9(y+4))) +(1/(9(2y-1)))`

Now,

`inty/((y+4)(2y-1)) dy`

=` int [(4/(9(y+4))) +(1/(9(2y-1)))]dy`

= `int (4/(9(y+4))) dy + int (1/(9(2y-1)))dy`

= `(4/9) ln(y+4) + (1/9) *(1/2)(ln(2y-1)) +c`

= `(4/9) ln(y+4) + (1/18)(ln(2y-1)) +c`

is  the solution

:)

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