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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 $\displaystyle sqrt(x4)$

September 5th, 2014, 11:53 AM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 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: $\displaystyle I=\sinh(2u)+2u+C$ Backsubstituting 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: $\displaystyle I=\frac{x}{2}\sqrt{x^2+4}+ 2\ln\left(x+\sqrt{x^2+4}\right)+C$ 
September 5th, 2014, 04:57 PM  #4  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, Married Math! Quote:
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 $ $\qquad =\;4\cdot\frac{1}{2}\big[\sec\theta\tan\theta + \ln\sec\theta + \tan\theta\big] + C$ Backsubstitute: $\:\tan\theta \,=\,\dfrac{x}{2},\;\sec\theta \,=\,\dfrac{\sqrt{x^2+4}}{2}$ $\qquad 2\left[\dfrac{\sqrt{x^2+4}}{2}\cdot\dfrac{x}{2} + \ln\left\dfrac{\sqrt{x^2+4}}{2} + \dfrac{x}{2}\right \right] + C$ $\qquad =\;2\left[\dfrac{x\sqrt{x^2+4}}{4} + \ln\left\dfrac{x+\sqrt{x^2+4}}{2}\right\right] + C$  
September 5th, 2014, 05:07 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,462 Thanks: 2499 Math Focus: Mainly analysis and algebra 
I'm sure you meant $x = 2\tan\theta$.
Last edited by v8archie; September 5th, 2014 at 05:12 PM. 
September 6th, 2014, 06:00 AM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs  
September 7th, 2014, 10:18 AM  #7 
Newbie Joined: Sep 2014 From: uk Posts: 15 Thanks: 2 
Thanks guys

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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