# Need Help With This Integral

• TheRedDevil18

## Homework Statement

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

∫x/(x^2+2x+2) dx

## 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

What's the derivative of the denominator? Is it similar to the numerator?

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 ?

I would start by completing the square in the denominator- $x^2+ 2x+ 2= x^2+ 2x+ 1+ 1= (x+1)^2+ 1$
$$\int \frac{x}{x^2+ 2x+ 2} dx= \int \frac{x}{(x+ 1)^2+ 1} dx$$

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

I would start by completing the square in the denominator- $x^2+ 2x+ 2= x^2+ 2x+ 1+ 1= (x+1)^2+ 1$
$$\int \frac{x}{x^2+ 2x+ 2} dx= \int \frac{x}{(x+ 1)^2+ 1} dx$$

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|

ln|√(x+1)^2+1| - arctan(x+1) + C

Correct ?

## Homework Statement

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

∫x/(x^2+2x+2) dx

## 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.

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 ?

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.

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.

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 ?