# How To Integrate 1/[sqrt (x^2 + 3x + 2)] dx?

How do you integrate ##\frac{1}{\sqrt{x^2 + 3x + 2}} dx##?

I had tried using ##u = x^2 + 3x + 2## and trigonometry substitution but failed.

Please give me some clues and hints.

Thank you

mentor note: moved from a non-homework to here hence no template.

## Answers and Replies

PeroK
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Complete the square.

jedishrfu
This is my attempt for the completing the square:

##x^2 + 3x + 2 = 0##
##x^2 + 3x = -2##
##x^2 + 3x + 2.25 = -2 + 2.25##
##x^2 + 3x + 2.25 = 0.25##
##(x + 1.5)^2 = 0.25##
##x + 1.5 = \sqrt{0.25}##
##x + 1.5 = ± 0.5##
##x_{1,2} = (± 0.5) - 1.5##
##x_1 = 0.5 - 1.5 = -1##
##x_2 = -0.5 - 1.5 = -2##

Is this correct?

What is the next step?

PeroK
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I think you've solved a quadratic equation there. You were only supposed to complete the square:
$$x^2 + 3x + 2 = (x+a)^2 - b^2$$
You need to find ##a## and ##b##.

I can't find the "completing the square" in my algebra book. Please tell me what page of below attached algebra book about completing the square?

https://www.sendspace.com/file/o2yqxe

The only completing the square I found in the above book is what I wrote in post #3.

PeroK
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It's what i put in post #4. Google if you want.

Getting rid of the linear term might be a better description.

mathman
Try ##y=x+1.5## so you have ##\frac{dy}{\sqrt{y^2-0.25}}##. Trig substitution may work.

Dr Transport
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factor the $x^2 + 3x +2$ into $(x+2)(x+1)$ and expand into partial fractions then use trig substitution.

Homework Helper
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factor the $x^2 + 3x +2$ into $(x+2)(x+1)$ and expand into partial fractions then use trig substitution.
I think the square root makes it so that partial fractions will not work, but trigonometric substitution after completing the square is straightforward. Edit: The trigonometric substitution is somewhat complicated. See also post 11.

Last edited:
mathman
Related ##\int\frac{dx}{\sqrt{x^2-1}}=ln(x+\sqrt{x^2-1})##. I don't see any obvious simple method.

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Related ##\int\frac{dx}{\sqrt{x^2-1}}=ln(x+\sqrt{x^2-1})##. I don't see any obvious simple method.
The trigonometric substitution ## x=\sec{\theta} ## will work. I used the word straightforward in post 9, but upon working through it in detail, I see it involves computing ## \int \sec{\theta} \, d \theta ## which involves a "trick", so it is somewhat complicated.

PeroK
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Related ##\int\frac{dx}{\sqrt{x^2-1}}=ln(x+\sqrt{x^2-1})##. I don't see any obvious simple method.
##x = \cosh u##

##\cosh^{-1}x = \ln(x + \sqrt{x^2 - 1})##

Mark44
Mentor
I can't find the "completing the square" in my algebra book. Please tell me what page of below attached algebra book about completing the square?
I looked at the link you posted, but I would have needed to download the book, and then search through it, which I don't care to do.

In your book, find the chapter or section that presents the Quadratic Formula. In that section of somewhere before it, they should talk about "Completing the Square." If your book lays out a derivation of the Quadratic Formula, it's done using the technique of completing the square.

PeroK
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I looked at the link you posted, but I would have needed to download the book, and then search through it, which I don't care to do.

In your book, find the chapter or section that presents the Quadratic Formula. In that section of somewhere before it, they should talk about "Completing the Square." If your book lays out a derivation of the Quadratic Formula, it's done using the technique of completing the square.
It's all over the Internet. Here, for example:

https://en.wikipedia.org/wiki/Completing_the_square

Grasshopper and Klystron
I can't find the "completing the square" in my algebra book. Please tell me what page of below attached algebra book about completing the square?

https://www.sendspace.com/file/o2yqxe

The only completing the square I found in the above book is what I wrote in post #3.
Hi! Disclaimer: I’m a student.

To complete the square, take the quantity x + half of the second term and square it. Then add the appropriate constant that keeps the polynomial the same.

For example:

##x^2 + bx + c = (x + \frac{b}{2} )^2 + c - \frac{b^2}{4}##

PeroK
Hey what do you do after completing the square? Factoring this is actually really easy. Even doing a partial fraction is cake, under the radical. But I can’t see how that helps. Unless... I’m hoping completing the square leaves it in a useful form. Hmm...

I think you can get a difference of squares under the radical, since ##\frac{1}{4}## is a perfect square.

Maybe trig substitution then? Once you get it in the form of ##\frac{1}{\sqrt{a^2 - b^2}}##

But in this particular case I think a is actually a(x), so I think you’d have to do sub and then a trig sub, maybe.

This is a toughy.

Edit - actually scratch that, cause I think only the form ##\frac{1}{\sqrt{1- x^2}}## would be useful for that.

Feel free to delete if this adds nothing. Fun problem.

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See posts 10, 11, and 12. The substitution of post 12 is perhaps the easiest.

Thanks Charles. The only reason I didn’t utilize that was because it felt like it came from an integral table. But the sec substitution is probably one I should know, so disregard this haha.

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Thanks Charles. The only reason I didn’t utilize that was because it felt like it came from an integral table. But the sec substitution is probably one I should know, so disregard this haha.
Yes, and to then do the complete calculation, you wind up integrating ## \int \sec{\theta} \, d \theta ##. That one involves a trick of multiplying numerator and denominator by ## \sec{\theta}+\tan{\theta} ## to get ## \int \frac{ d (\sec{\theta}+\tan{\theta})}{\sec{\theta}+\tan{\theta}}=\ln|\sec{\theta}+\tan{\theta} | ##. It's basically a trick, and without seeing it, you wouldn't be expected to be able to solve the integral ## \int \sec{\theta} \, d \theta ##.

Grasshopper
I finally found how to completing the square of ##x^2 + 3x + 2##, and here is the result:

##(x + 1.5)^2 - (\sqrt{0.25})^2##

Am I correct?

If it correct, then

##\int \frac{1}{\sqrt{x^2 + 3x + 2}} dx##
## = \int \frac{1}{\sqrt{(x + 1.5)^2 - (\sqrt{0.25})^2}} dx##

Let ##u = x + 1.5##, then ##du = dx##

## = \int \frac{du}{\sqrt{u^2 - (\sqrt{0.25})^2}}##

What next?

Last edited:
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You might try computing ## \sqrt{0.25} ##. The result is simple.

You might try computing ## \sqrt{0.25} ##. The result is simple.

##\sqrt{0.25} = 0.5##

so

## = \int \frac{du}{\sqrt{u^2 - (\sqrt{0.25})^2}}##
## = \int \frac{du}{\sqrt{u^2 - (0.5)^2}}##

What next?

Homework Helper
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See posts 10,11, and 12. You also need the substitution ## u=x+1.5 ##