Need Help With This Integral

  • #1

Homework Statement



Hi guys, I need help on how to go about solving this integral

∫x/(x^2+2x+2) dx

Homework Equations





The Attempt at a Solution



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 please help me, thanks
 

Answers and Replies

  • #2
SteamKing
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What's the derivative of the denominator? Is it similar to the numerator?
 
  • #3
What's the derivative of the denominator? Is it similar to the numerator?

It's 2x+2, do you mean I should use u substitution ?
 
  • #4
HallsofIvy
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I would start by completing the square in the denominator- [itex]x^2+ 2x+ 2= x^2+ 2x+ 1+ 1= (x+1)^2+ 1[/itex]
[tex]\int \frac{x}{x^2+ 2x+ 2} dx= \int \frac{x}{(x+ 1)^2+ 1} dx[/tex]

Now let u= x+ 1 so that du= dx and x= u- 1.
 
  • #5
I would start by completing the square in the denominator- [itex]x^2+ 2x+ 2= x^2+ 2x+ 1+ 1= (x+1)^2+ 1[/itex]
[tex]\int \frac{x}{x^2+ 2x+ 2} dx= \int \frac{x}{(x+ 1)^2+ 1} dx[/tex]

Now let u= x+ 1 so that du= dx and x= u- 1.

Thanks, this is what I have tried, correct me if I'm wrong

by completing the square and u sub

∫(u-1)/(u^2+1) du

= ∫ u/(u^2+1) - ∫1/(u^2+1)

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

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

plugging back in I get,

∫(tan(t)*sec(t)^2)/(sec^2(t)) dt...................tan(t)^2 + 1 = sec(t)^2

= ∫tan(t) dt
= ln |sec(t)|

Using the triangle thing, I get

ln|√u^2+1|

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

ln|√(x+1)^2+1|

Final answer

ln|√(x+1)^2+1| - arctan(x+1) + C

Correct ?
 
  • #6
LCKurtz
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Homework Statement



Hi guys, I need help on how to go about solving this integral

∫x/(x^2+2x+2) dx

Homework Equations





The Attempt at a Solution



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 please help me, thanks

Here's what your prof was suggesting:$$
\int\frac x {x^2+2x+2}~dx = \frac 1 2 \int\frac {2x}{x^2+2x+2}~dx
=\frac 1 2 \int\frac{(2x+2)-2}{x^2+2x+2}~dx =\frac 1 2\int\frac{2x+2}{x^2+2x+2}~dx
-\int \frac 1 {x^2+2x+2}$$Now the first integral is set up for the natural u substitution and the second is ready to complete the square as others have suggested. Personally, I would just complete the square in the first place as Halls has suggested. Your prof's method is sometimes useful.
 
  • #7
Here's what your prof was suggesting:$$
\int\frac x {x^2+2x+2}~dx = \frac 1 2 \int\frac {2x}{x^2+2x+2}~dx
=\frac 1 2 \int\frac{(2x+2)-2}{x^2+2x+2}~dx =\frac 1 2\int\frac{2x+2}{x^2+2x+2}~dx
-\int \frac 1 {x^2+2x+2}$$Now the first integral is set up for the natural u substitution and the second is ready to complete the square as others have suggested. Personally, I would just complete the square in the first place as Halls has suggested. Your prof's method is sometimes useful.

Thanks, I'm a bit confused though

Where did the 1/2 come from ?

and, the derivative of x^2+2x+2 = 2x+2, so where did the 2x come from ?
 
  • #8
BiGyElLoWhAt
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the 1/2 is a "multiply by 1" thing. Multiply x by 2 and then the whole fraction (effectively the x again) by 1/2.
 
  • #9
LCKurtz
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Thanks, I'm a bit confused though

Where did the 1/2 come from ?

and, the derivative of x^2+2x+2 = 2x+2, so where did the 2x come from ?

You are trying to build a ##2x+2## in the numerator so you start by writing ##x = \frac 1 2\cdot 2x##. Then you add and subtract ##2##. That's what your prof meant by forcing the derivative of the denominator into the numerator.
 
  • #10
the 1/2 is a "multiply by 1" thing. Multiply x by 2 and then the whole fraction (effectively the x again) by 1/2.

Oh, yes I see, but isn't that suppose to be 2x+2 (referring to integral with the 2x) as that's the derivative of the denominator ?
 

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