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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.

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In summary, you integrate ##\frac{1}{\sqrt{x^2 + 3x + 2}} dx## by using trigonometric substitution and then factoring to get a difference of squares.

- #1

- 169

- 9

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.

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- #2

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Complete the square.

- #3

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##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?

- #4

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$$x^2 + 3x + 2 = (x+a)^2 - b^2$$

You need to find ##a## and ##b##.

- #5

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

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

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

- #7

Science Advisor

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

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- #9

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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.Dr Transport said:

Last edited:

- #10

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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.

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

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

- #13

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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.askor said: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?

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.

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It's all over the Internet. Here, for example:Mark44 said: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.

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

- #16

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Hi! Disclaimer: I’m a student.askor said:

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

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

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}##

- #17

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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.

- #19

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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 said:

- #21

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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?

##(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:

- #22

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

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Charles Link said: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?

- #24

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

- #25

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The simplest next step is essentially given by @PeroK in post 12: ## v=.5 \cosh{u} ##, but you need to show more effort here. PF rules are that the Homework Helper cannot supply the solution, and that is nearly happening here. See if you can solve it from here.askor said:What next?

- #26

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There's a less tricky way to integrate that.Charles Link said: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 ##.

$$\int \sec \theta\,d\theta = \int \frac 1{\cos \theta}\,d\theta = \int \frac {\cos\theta}{\cos^2 \theta}\,d\theta = \int \frac {\cos\theta}{1-\sin^2 \theta}\,d\theta$$ then use ##u=\sin\theta##.

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Charles Link said:The simplest next step is essentially given by @PeroK in post 12: ## v=.5 \cosh{u} ##, but you need to show more effort here. PF rules are that the Homework Helper cannot supply the solution, and that is nearly happening here. See if you can solve it from here.

Why use hyperbolic cos instead of standard/regular cos?

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vela said:There's a less tricky way to integrate that.

$$\int \sec \theta\,d\theta = \int \frac 1{\cos \theta}\,d\theta = \int \frac {\cos\theta}{\cos^2 \theta}\,d\theta = \int \frac {\cos\theta}{1-\sin^2 \theta}\,d\theta$$ then use ##u=\sin\theta##.

You said use ##u = \sin \theta##, then the form will be changing like this:

##\int \frac{\cos \theta}{1 - u^2} d\theta##

What should I do with the ##\cos \theta##?

- #29

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Because you can. Try it both ways and see why you might choose one over the other.askor said:Why use hyperbolic cos instead of standard/regular cos?

Ask again after you finish rewriting everything in terms of ##u##.askor said:What should I do with the ##\cos \theta##?

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vela said:Ask again after you finish rewriting everything in terms of ##u##.

I see.

If ##u = \sin \theta## then ##du = \cos \theta \, d \theta##

Then the form will be like this:

##\int \frac{du}{1 - u^2}##

What should I do next?

My brain is fatigue.

- #31

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Take a break and try again later with fresh eyes.

- #32

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Using ##\cosh## is about a hundred million time better. If you'll excuse the hyperbole!vela said:There's a less tricky way to integrate that.

$$\int \sec \theta\,d\theta = \int \frac 1{\cos \theta}\,d\theta = \int \frac {\cos\theta}{\cos^2 \theta}\,d\theta = \int \frac {\cos\theta}{1-\sin^2 \theta}\,d\theta$$ then use ##u=\sin\theta##.

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PeroK said:Using ##\cosh## is about a hundred million time better. If you'll excuse the hyperbole!

How do you use ##\cosh##? Please show me an example.

- #34

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Have the hyperbolic trig functions been presented to you, yet? If not, an example probably won't make much sense.askor said:How do you use ##\cosh##? Please show me an example.

- #35

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PeroK said:If you'll excuse the hyperbole!

I saw what you did there.

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