Difficult integration by parts

[hoeranski]

Homework Statement

Calculate the following integral with partial integration:

\int\frac{1 - x}{(x^2 + 2x + 3)^2}dx

The Attempt at a Solution

I guess you need to write the integral in easier chunks but I still fail every time.

 

[sjb-2812]

Can you write this as partial fractions? or make use of the fact that the bottom can be factorised?

 

[verty]

I haven't tried this but letting u = 1/(x^2 + 2x + 3) may get you somewhere.

Sorry, I've checked this now and it doesn't work. Another angle of attack, probably not the best method though, is u = x + b/2a = x + 1.

 

[verty]

Making the substitution I suggested (u = x + 1), one gets this:

$\int{\frac{2-u}{(2+u^2)^2}} du$

Then, one has a choice of substitutions (tan, cot, sinh). By inspection, sinh won't work. We then have denominator cosh^3, not nice. That leaves tan and cot although it seems tan is never a worse substitution than cot. I hold out hope but... use tan and...

Best of luck.

PS. Given the nature of this question, I can offer no further help. A difficult question is meant to be difficult.

 

[HallsofIvy]

With a quadratic that cannot be factored (using real numbers) in the denominator, the standard step is to complete the square: x^3+ 2x+ 3= x^2+ 2x+ 1+ 2= (x+ 1)^2+ 2 so you can write the integral as \int\frac{1- x}{((x+1)^2+ 2)^3}dx.

Now let t= x+ 1 so that dx= dt and 1- x= 2-(x+1)= 2- t. With that substitution, the integral becomes
\int\frac{2- t}{(t^2+ 2)^2}dt= 2\int\frac{dt}{(t^2+ 2)^2}- \frac{t}{(t^2+ 2)^2}dt
For the first integral try t= \sqrt{2}tan(\theta) and for the second, u= t^2+ 2.

 

[Millennial]

Another way to tackle the integral could be to somehow create the derivative of the base for the exponent in the denominator in the numerator. Then, you could use substitution.

 

[lurflurf]

^That is pretty much what they did
$$\frac{1-x}{(x^2+2x+3)^2}=\frac{2-(x^2+2x+3)^\prime /2}{(x^2+2x+3)^2}$$
then we need to split it into two parts otherwise we need to use complex numbers