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September 5th, 2014, 11:05 AM   #1
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How can I solve this Integration sqrt(x^2+4)

How can I solve this Integration with steps

$\displaystyle sqrt (x^2+4)$

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September 5th, 2014, 11:32 AM   #2
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or $\displaystyle sqrt(x-4)$
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September 5th, 2014, 11:53 AM   #3
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We are given:

$\displaystyle I=\int \sqrt{x^2+4}\,dx$

When I see the sum of two squares under a radical in the integrand, I think of the hyperbolic identity:

$\displaystyle \cosh^2(u)=1+\sinh^2(u)$

And so if we let:

$\displaystyle x=2\sinh(u)\,\therefore\,dx=2\cosh(u)\,du$

We obtain:

$\displaystyle I=4\int \cosh^2(u)\,du$

Now, using the identity:

$\displaystyle \cosh^2(u)=\frac{\cosh(2u)+1}{2}$

We have:

$\displaystyle I=\int \cosh(2u)+1\,d(2u)$

And so integrating term by term, there results:

$\displaystyle I=\sinh(2u)+2u+C$

Back-substituting for $u$, we get:

$\displaystyle I=\frac{x}{2}\sqrt{x^2+4}+ 2\text{arsinh}\left(\frac{x}{2}\right)+C$

If we convert the inverse hyperbolic function to a logarithm, we may state:

$\displaystyle I=\frac{x}{2}\sqrt{x^2+4}+ 2\ln\left(x+\sqrt{x^2+4}\right)+C$
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September 5th, 2014, 04:57 PM   #4
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Hello, Married Math!

Quote:
$\displaystyle\int \sqrt {x^2+4}\,dx $

Trig Substitution . . .

Let $x \,=\,2\sec\theta \quad\Rightarrow\quad dx \,=\,2\sec^2\theta\,d\theta \quad\Rightarrow\quad \sqrt{x^2+4} \,=\,2\sec\theta$

Substitute: $\:\displaystyle\int(2\sec\theta)(2\sec^2\theta\,d \theta) \;=\;4\int\sec^3\theta\,d\theta $

$\qquad =\;4\cdot\frac{1}{2}\big[\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\big] + C$

Back-substitute: $\:\tan\theta \,=\,\dfrac{x}{2},\;\sec\theta \,=\,\dfrac{\sqrt{x^2+4}}{2}$

$\qquad 2\left[\dfrac{\sqrt{x^2+4}}{2}\cdot\dfrac{x}{2} + \ln\left|\dfrac{\sqrt{x^2+4}}{2} + \dfrac{x}{2}\right| \right] + C$

$\qquad =\;2\left[\dfrac{x\sqrt{x^2+4}}{4} + \ln\left|\dfrac{x+\sqrt{x^2+4}}{2}\right|\right] + C$

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September 5th, 2014, 05:07 PM   #5
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I'm sure you meant $x = 2\tan\theta$.
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Last edited by v8archie; September 5th, 2014 at 05:12 PM.
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September 6th, 2014, 06:00 AM   #6
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Quote:
Originally Posted by Married Math View Post
or $\displaystyle sqrt(x-4)$
$\displaystyle \int\sqrt{x-4}\,dx=\frac{2}{3}(x-4)^{\frac{3}{2}}+C$
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September 7th, 2014, 10:18 AM   #7
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Thanks guys
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September 19th, 2014, 07:56 AM   #8
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It's very hard
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