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

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and even what @PeroK presented in post 12 takes a little algebra to compute, starting with ## x=\cosh{u}=\frac{e^u+e^{-u}}{2} ##.

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I only know the very basic properties of hyperbolic function such as:

##\sinh x = \frac{e^x - e^{-x}}{2}##

and

##\cosh x = \frac{e^x + e^{-x}}{2}##

But I don't know how to use it in technique of integration.

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I only know the very basic properties of hyperbolic function such as:

##\sinh x = \frac{e^x - e^{-x}}{2}##

and

##\cosh x = \frac{e^x + e^{-x}}{2}##

But I don't know how to use it in technique of integration.
Using the above definitions what are the derivatives of ##\sinh x## and ##\cosh x##?

What is ##\cosh^2 x## in terms of ##\sinh^2 x##?

In terms of integration, you use them the same way you use the trig functions, by substitution. E.g.:
$$x = \cosh u, \ \ dx = \frac{d}{du} (\cosh u) du$$

I only know the very basic properties of hyperbolic function such as:

##\sinh x = \frac{e^x - e^{-x}}{2}##

and

##\cosh x = \frac{e^x + e^{-x}}{2}##

But I don't know how to use it in technique of integration.
I think finding ##dx## helps so you need to find the derivative of ##\cosh x##. I learned it from a table of derivatives and integrals, which contained also ##\cosh x##.

octopus26
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.
Wolfram Alpha

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##\int \frac{du}{1 - u^2}##
The hyperbolic trig approach is neater, but to proceed with the above form use partial fractions.

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The hyperbolic trig approach is neater, but to proceed with the above form use partial fractions.
With you have there, why couldn’t a second substitution be made, using ##1 - sin^2w = cos^2w##?

I’m sure I’m missing something.

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With you have there, why couldn’t a second substitution be made, using ##1 - sin^2w = cos^2w##?

I’m sure I’m missing something.
That would be going back to what we had earlier, integrating sec.
Do you see how to solve it using partial fractions?

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That would be going back to what we had earlier, integrating sec.
Do you see how to solve it using partial fractions?
Wait, lol sorry I see now. It was late I was not thinking clearly. But yeah this is one of the easier partial fractions.

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Wait, lol sorry I see now. It was late I was not thinking clearly. But yeah this is one of the easier partial fractions.
what is the final solution? did you really get it?