I have a problem solving this partial fraction... Please somebody help me out...

Integrate the following partial fraction:

2X + 2

__________________ dx

(x^2 + 1 ) (x + 1)^3

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- Apr 19th 2005, 03:11 PMchinoPartial Fractions HELP PLEASE!!!
I have a problem solving this partial fraction... Please somebody help me out...

Integrate the following partial fraction:

2X + 2

__________________ dx

(x^2 + 1 ) (x + 1)^3 - Apr 19th 2005, 06:33 PMpaultwang
Perform a partial fraction expansion and integrate each term individually. Where are you stuck?

- Apr 19th 2005, 07:17 PMMath HelpHow about this?
__2x+2 dx__(x^2+1)(x+1)^3

Couldn't you do this:

.......x+1

2-------------------dx

(x^2+1)(X+1)^3

.........1

2-------------------dx

(x^2+1)(X+1)^2

then 2 * (x^2+1)^-1 * (x+1)^-2 dx

Then I assume multiply it out an integrate normally. - Apr 20th 2005, 04:47 AMticbol
2X + 2

__________________ dx

(x^2 + 1 ) (x + 1)^3

Rewriting that in one line only,

(2x+2)/[(x^2 +1)(x+1)^3] dx

Simplifying it because there is a factor (x+1) that is common to both numerator and denominator,

= 2(x+1)/[(x^2 +1)(x+1)^3] dx

= 2/[(x^2 +1)(x+1)^2] dx

You want to decompose 2/[(x^2 +1)(x+1)^2] into partial fractions.

Okay.

I hope you know the process so that I don't have to explain it. (If not, then ask me.)

2/[(x^2 +1)(x+1)^2] = (Ax +B)/(x^2 +1) +C/(x+1) +D/(x+1)^2 ....(1)

Multiply both sides by [(x^2 +1)(x+1)^2],

2 = (Ax +B)(x+1)^2 +C(x^2 +1)(x+1) +D(x^2 +1) ....(2)

When x=(-1), plug that into (2),

2 = (A(-1) +B)(-1+1)^2 +C(x^2 +1)(-1+1) +D((-1)^2 +1)

2 = 0 +0 +D(2)

So,

D = 2/2 = 1 ....***

When x=0, plug that into (2), ...and plug also D=1,

2 = (A(0) +B)(0+1)^2 +C(0^2 +1)(0+1) +1(0^2 +1)

2 = (B)(1) +C(1)(1) +1(1)

2 = B +C +1

2 -1 = B +C

1 = B +C ....(3)

When x=1, plug that into (2), ...and plug also D=1,

2 = (A(1) +B)(1+1)^2 +C(1^2 +1)(1+1) +1(1^2 +1)

2 = (A+B)(2)^2 +C(2)(2) +1(2)

2 = (A+B)(4) +C(4) +2

2 -2 = 4(A+B) +4C

0 = 4(A+B) +4C

0 = (A+B) +C

0 = A +B +C ...(4)

Eq.(4) minus Eq.(3),

-1 = A ....***

When x=2, plug that into (2), ...and plug also D=1 and A = -1,

2 = (-1(2) +B)(2+1)^2 +C(2^2 +1)(2+1) +1(2^2 +1)

2 = (-2 +B)(3)^2 +C(5)(3) +1(5)

2 = (-2 +B)(9) +15C +5

2 = -18 +9B +15C +5

2 +18 -5 = 9B +15C

15 = 9B +15C

Divide both sides by 3,

5 = 3B +5C ....(5)

From (3), C = 1-B, plug that into (5),

5 = 3B +5(1-B)

5 = 3B +5 -5B

5 -5 = -2B

B = 0 ....***

Susbtitute that in (3),

C = 1 -0 = 1 ...***

So we found

A = -1

B = 0

C = 1

D = 1

Hence,

2/[(x^2 +1)(x+1)^2] = (Ax +B)/(x^2 +1) +C/(x+1) +D/(x+1)^2 ....(1)

2/[(x^2 +1)(x+1)^2] = (-1x +0)/(x^2 +1) +1/(x+1) +1/(x+1)^2

2/[(x^2 +1)(x+1)^2] = -x/(x^2 +1) +1/(x+1) +1/(x+1)^2 .....******

There, it is decomposed.

------------------

So,

INT.[(2x +2)/[(x^2 +1)(x+1)^3]]dx

= INT.[2 /[x^2 +1)(x+1)^2]]dx

= INT.[-x/(x^2 +1) +1/(x+1) +1/(x+1)^2]dx

= -(1/2)ln(x^2 +1) +ln(x+1) -1/(x+1) +C ....answer. - Apr 20th 2005, 01:08 PMchino
Thank you guys for the help. This is one of the partial fraction problems where one little mistake will cost you millions of headaches.