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How to solve the definite integral of (3x^4 +22x^3 +57x^2 +69x +36) / (x^3 +6x^2 +9x)

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smartguy2323 | Honors

Posted May 31, 2013 at 8:38 PM via web

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How to solve the definite integral of (3x^4 +22x^3 +57x^2 +69x +36) / (x^3 +6x^2 +9x)

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crmhaske | College Teacher | (Level 3) Associate Educator

Posted June 1, 2013 at 12:02 AM (Answer #1)

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As no interval has been given, only the indefinite integral can be calcuated.

Factor the expression:

`f(x)=(3x^4 +22x^3 +57x^2 +69x +36)/(x^3 +6x^2 +9x)`

`=(3x^4 +22x^3 +57x^2 +69x +36)/(x(x^2+6x+9))`

`=(3x^4 +22x^3 +57x^2 +69x +36)/(x(x+3)^2)`

Long division:

                   3x^2 +  4x + 6
x^2+6x+9 | 3x^4 + 22x^3 + 57x^2 + 69x +36
                   3x^4 + 18x^3 + 27x^2
                                 4x^3 + 30x^2 + 69x
                                 4x^3 + 24x^2 + 36x
                                               6x^2 + 33x + 36
                                               6x^2 + 36x + 54
                                                           -3x - 18

`f(x)=(3x^2+4x+6)/x - (3x+18)/(x(x+3)^2)`

Separate terms:

`f(x)=3x+4+6/x- (3x+18)/(x(x+3)^2)`

Use partial fractions:

`a/x+b/(x+3)+c/(x+3)^2=(3x+18)/(x(x+3)^2)`

`a(x+3)^2+bx(x+3)+cx=3x+18`

`ax^2+6xa+9a+bx^2+3bx+cx`

`(a+b)x^2+(6a+3b+c)x+9a=3x+18`

`a+b=0`

`6a+3b+c = 3`

`9a = 18 -gt a=2`

`2+b=0-gtb=-2`

`6(2)+3(-2)+c=3`

`c=3+6-12=-3`

`2/x-2/(x+3)-3/(x+3)^2=(3x+18)/(x(x+3)^2)`

Now we have:

`f(x)=3x+4+6/x-2/x+2/(x+3)+3/(x+3)^2`

`=3x+4+4/x+2/(x+3)+3/(x+3)^2`

Now we can integrate the function:

`intf(x)dx=int3xdx+int4dx+int4/xdx+int2/(x+3)dx+int3/(x+3)^2dx`

`=3/2x^2+4x+4ln(x)+2ln(x+3)-3/(x+3)+C`

 

Sources:

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