Need Help With This Integral

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Scremin34Egl

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Hi guys, I need help trying to solve this integral

integral: x/(x^2+2x+2) dx

My math professor said something about "force the derivative of denominator to occur at top and compensate" but I have no idea what he means. Can somebody help me, thanks
 
Scremin34Egl said:
integral: x/(x^2+2x+2) dx

My math professor said something about "force the derivative of denominator to occur at top and compensate" but I have no idea what he means.
Click to expand...
What is the derivative of the denominator? How does this compare with the numerator? Is there some "fudge factor" that you can introduce to create a relationship between the two?

In other words, please reply showing what you have tried and how far you have gotten in attempting to follow the instructions you've been given. Thank you! ;)
 
just to help notation, let
p = (x^2+2x+2)
then
p' = 2 x + 2

since x = ( 2x + 2 - 2) / 2 we have x = 1/2 p' - 1. Thus
I(x/p) = 1/2 I(p'/p) - I(1/p)
where I(f) indicates the integral of f. I(p'/p) [involves ln function] and I(1/p) [involves arctangent function] are standard integrals which you should know or be able to work out quickly although I must admit, it has been a while for me seeing I(1/p) and I had to look it up.
 
Ok, this is what I have tried

By completing the square and u sub

integral: (u-1)/(u^2+1) du

= integral: (u)/(u^2+1) - integral: 1/(u^2+1)

For the first integral I used a trig substitution, like this

u = tan(t)
du = sect(t)^2 dt

plugging back in, I get

integral: (tan(t)*sec(t)^2)/sec(t)^2 dt

= integral: tan(t) dt
= ln |sec(t)|

Using the triangle thing, I get

ln|sqrt(u^2+1)|

and, u= x+1, so plugging back in

ln |sqrt(x+1)^2+1|

Final answer

ln |sqrt(x+1)^2+1| - arctan(x+1) + C

Correct ?
 
Scremin34Egl said:
Ok, this is what I have tried

By completing the square and u sub

integral: (u-1)/(u^2+1) du

= integral: (u)/(u^2+1) - integral: 1/(u^2+1)

For the first integral I used a trig substitution, like this

u = tan(t)
du = sect(t)^2 dt

plugging back in, I get

integral: (tan(t)*sec(t)^2)/sec(t)^2 dt

= integral: tan(t) dt
= ln |sec(t)|

Using the triangle thing, I get

ln|sqrt(u^2+1)|

and, u= x+1, so plugging back in

ln |sqrt(x+1)^2+1|

Final answer

ln |sqrt(x+1)^2+1| - arctan(x+1) + C

Correct ?
Click to expand...

Differentiate your answer - if you get back the original expression - you are most probably correct. Try it....
 
Scremin34Egl said:
Ok, this is what I have tried ...
Correct ?
Click to expand...

Yes, to a point. But that is from someone who learned the vast majority of their math and got their degree in mathematics a long time ago and still likes to play around.

Continuing my post above and referring to your math professors comment "force the derivative of denominator to occur at top and compensate", that is what was done by writing
x / (x^2+2x+2) = x / p,

x = (1/2 dp/dx - 1) = 1/2 p' -1,

and obtaining the two standard intervals [standard after making the other substitution as you did, u = x+1]

I(p'/p) = ln(p)
and
I(1/p) = arctan(x+1)

Thus we have
I(x / (x^2+2x+2)) = 1/2 ln (x^2+2x+2) - arctan(x+1) + C

which is equivalent to what you have, i.e.
ln (sqrt(x^2+2x+2)) = ln ((x^2+2x+2)^(1/2)) = 1/2 ln (x^2+2x+2) = 1/2 ln(p)

Note that there is no absolute value of p. In this particular case it makes no difference since p is positive for all real valued x.

BTW: Nice work
 
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Ishuda said:
Yes, to a point. But that is from someone who learned the vast majority of their math and got their degree in mathematics a long time ago and still likes to play around.

Continuing my post above and referring to your math professors comment "force the derivative of denominator to occur at top and compensate", that is what was done by writing
x / (x^2+2x+2) = x / p,

x = (1/2 dp/dx - 1) = 1/2 p' -1,

and obtaining the two standard intervals [standard after making the other substitution as you did, u = x+1]

I(p'/p) = ln(p)
and
I(1/p) = arctan(x+1)

Thus we have
I(x / (x^2+2x+2)) = 1/2 ln (x^2+2x+2) - arctan(x+1) + C

which is equivalent to what you have, i.e.
ln (sqrt(x^2+2x+2)) = ln ((x^2+2x+2)^(1/2)) = 1/2 ln (x^2+2x+2) = 1/2 ln(p)

Note that there is no absolute value of p. In this particular case it makes no difference since p is positive for all real valued x.

BTW: Nice work
Click to expand...

Thank you, I will look into that method and get back if I have any problems :)
 
Scremin34Egl said:
Hi guys, I need help trying to solve this integral

integral: x/(x^2+2x+2) dx

My math professor said something about "force the derivative of denominator to occur at top and compensate" but I have no idea what he means. Can somebody help me, thanks
Click to expand...

Same as Ishuda's method - but slightly different steps:

\(\displaystyle \frac{d}{dx}[x^2+2x+2] \ = \ 2x + 2\) .... this (plus a constant) is what we want as numerator.

Then,

The given numerator \(\displaystyle x \ = \ \frac{1}{2}[2x+2] \ - \ 1 \)

\(\displaystyle \displaystyle{\int \frac{x}{x^2+2x+2}dx \ = \ \int \dfrac{\frac{1}{2}[2x+2] \ - \ 1}{x^2+2x+2} dx}\)

\(\displaystyle \displaystyle{\int \frac{x}{x^2+2x+2}dx \ = \ \frac{1}{2}\int \dfrac{2x+2}{x^2+2x+2} dx \ - \ \int \frac{1}{(x+1)^2+1^2} dx}\)

Now these are reduced to forms of standard integrals.
 
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