Question

Can someone help with this integration? We just learned integrations by manipulation of quadratic expressions.

∫(1/((x^2 + 4x + 5)^2))dx

Answer 1

This may help with visualizing it.

\[∫\frac{ dx }{ (x^2 + 4x + 5)^2 }\]

Answer 2

by manipulation of quadratics eh .. never heard of that method. Can you explain it?

Answer 3

the wolf gives a nice result, so the process would seem to be relatively easy to follow; im just not aware of the method

Answer 4

Essentially, from how I understood it, you get the denominator into a square form by adding constants inside and outside of the expression:

((x^2 + 4x + 4) - 4 + 5)^2
((x + 2)^2 + 1)^2

I thought you could use u-substitution here, but it doesn't get me anywhere (at least I don't think it does)...

u = x + 2
du = dx

--> ((u^2) + 1)^2

So now you'd have...

\[∫\frac{ du }{ (u^2 + 1)^2 }\]
It's at this step that I'm confused as to what to do next. I have a feeling that I have to do some wacky inverse trig substitutions, but I can't identify what exactly I have to do...

Answer 5

from there it would be partial decomp ...
\[\frac{ 1 }{ (u^2 + 1)^2 }=\frac{Au+B}{u^2+1}+\frac{Cu+D}{(u^2+1)^2}\]

Answer 6

when u=0

1 = B + D


when u = 1

1 = (A + B )/2 + (C+D)/4
4 = 2A + 2B + C + D

if we develop 4 equations in 4 unknowns, we can solve for ABCD

Answer 7

i did bad when u=1 .. i ignored the denominator on the left :(

hard part is keeping track is all

Answer 8

when u = 1

1/4 = (A + B )/2 + (C+D)/4

1 = 2A + 2B + C + D

thats better

Answer 9

when u = -1

1/4 = (-A + B )/2 + (-C+D)/4

1 = -2A + 2B - C + D


for simplicity, let u = 2

1/25 = (2A + B )/5 + (2C+D)/25

1 = 10A + 5B + 2C + D



0A + 1B + 0C+ 1D = 1
2A + 2B + 1C + 1D = 1
-2A + 2B - 1C + 1D = 1
10A + 5B + 2C + 1D = 1


well that didnt turn out as well as i was hoping it would :/ ABC = 0, D = 1

Answer 10

I was actually gonna say... I think I managed to work out a way to fit in a u-substitution. Could you say that, if going back to...

\[∫\frac{ du }{ (u^2 + 1)^2 }\]
u = tan ø
du = sec^2(ø) dø

...and then work from there?

Answer 11

its worth a shot

Answer 12

id love to stay and work over it but my times up, going home for the night ....

Answer 13

NP, thanks anyway! I think I can work my way through it eventually...

Answer 14

Yes, you're right about the trig sub:
\[\int\frac{ du }{ (u^2 + 1)^2 }\]
\(u=\tan t\), so \(du=\sec^2t ~dt\):
\[\int\frac{ \sec^2t }{ (\tan^2t + 1)^2 }~dt\\
\int\frac{ \sec^2t }{ (\sec^2t)^2 }~dt\\
\int\frac{dt}{ \sec^2t}\\
\int \cos^2t~dt\]

Answer 15

@amistre64, not sure, but i think the partial fraction decomp will result in the same function, i.e. A=B=C=0 and D=1.

Answer 16

From ∫cos^2(t) dt...

\[\frac{ 1 }{ 2 }∫(1 + \cos 2\theta) d \theta \]
= \[\frac{ 1 }{ 2 }∫d \theta + \frac{ 1 }{ 2 }∫ \cos 2\theta d \theta\]
= \[\frac{ \theta }{ 2 } + \frac{ \sin 2\theta }{ 4 } + C\]
And then just convert theta Ø back into terms of X, and that's it?

Answer 17

Yep, that's correct

Answer 18

yeah, when I rrefed the matrix i was horrified!! lol