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 September 5th, 2014, 11:05 AM #1 Newbie   Joined: Sep 2014 From: uk Posts: 15 Thanks: 2 How can I solve this Integration sqrt(x^2+4) How can I solve this Integration with steps $\displaystyle sqrt (x^2+4)$ Thanks
 September 5th, 2014, 11:32 AM #2 Newbie   Joined: Sep 2014 From: uk Posts: 15 Thanks: 2 or $\displaystyle sqrt(x-4)$
 September 5th, 2014, 11:53 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs 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: $\displaystyle I=\frac{x}{2}\sqrt{x^2+4}+ 2\ln\left(x+\sqrt{x^2+4}\right)+C$ Thanks from topsquark and Married Math
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: $\:\tan\theta \,=\,\dfrac{x}{2},\;\sec\theta \,=\,\dfrac{\sqrt{x^2+4}}{2}$

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

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

 September 5th, 2014, 05:07 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,341 Thanks: 2463 Math Focus: Mainly analysis and algebra I'm sure you meant $x = 2\tan\theta$. Thanks from soroban and Married Math Last edited by v8archie; September 5th, 2014 at 05:12 PM.
September 6th, 2014, 06:00 AM   #6
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Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,193
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Math Focus: Calculus/ODEs
Quote:
 Originally Posted by Married Math or $\displaystyle sqrt(x-4)$
$\displaystyle \int\sqrt{x-4}\,dx=\frac{2}{3}(x-4)^{\frac{3}{2}}+C$

 September 7th, 2014, 10:18 AM #7 Newbie   Joined: Sep 2014 From: uk Posts: 15 Thanks: 2 Thanks guys Thanks from MarkFL
 September 19th, 2014, 07:56 AM #8 Newbie   Joined: Sep 2014 From: uk Posts: 15 Thanks: 2 It's very hard

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